Fundamentals Of Image Processing
Fundamentals of Image Processing
hany.farid@dartmouth.edu
http://www.cs.dartmouth.edu/~farid
0. Mathematical Foundations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
0.1: Vectors
0.2: Matrices
0.3: Vector Spaces
0.4: Basis
0.5: Inner Products and Projections [*]
0.6: Linear Transforms [*]
1. Discrete-Time Signals and Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.1: Discrete-Time Signals
1.2: Discrete-Time Systems
1.3: Linear Time-Invariant Systems
2. Linear Time-Invariant Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.1: Space: Convolution Sum
2.2: Frequency: Fourier Transform
3. Sampling: Continuous to Discrete (and back) . . . . . . . . . . . . . . . . . . . . . . . . . 29
3.1: Continuous to Discrete: Space
3.2: Continuous to Discrete: Frequency
3.3: Discrete to Continuous
4. Digital Filter Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
4.1: Choosing a Frequency Response
4.2: Frequency Sampling
4.3: Least-Squares
4.4: Weighted Least-Squares
5. Photons to Pixels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
5.1: Pinhole Camera
5.2: Lenses
5.3: CCD
6. Point-Wise Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
6.1: Lookup Table
6.2: Brightness/Contrast
6.3: Gamma Correction
6.4: Quantize/Threshold
6.5: Histogram Equalize
7. Linear Filtering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
7.1: Convolution
7.2: Derivative Filters
7.3: Steerable Filters
7.4: Edge Detection
7.5: Wiener Filter
8. Non-Linear Filtering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
8.1: Median Filter
8.2: Dithering
9. Multi-Scale Transforms [*] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
10. Motion Estimation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
10.1: Differential Motion
10.2: Differential Stereo
11. Useful Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
11.1: Expectation/Maximization
11.2: Principal Component Analysis [*]
11.3: Independent Component Analysis [*]
[*] In progress
0. Mathematical Foundations
0.1 Vectors
0.1 Vectors
0.2 Matrices
From the preface of Linear Algebra and its Applications:
0.3 Vector Spaces
“Linear algebra is a fantastic subject. On the one hand
0.4 Basis
it is clean and beautiful.” – Gilbert Strang
0.5 Inner Products
This wonderful branch of mathematics is both beautiful and use-
and
ful. It is the cornerstone upon which signal and image processing
Projections
is built. This short chapter can not be a comprehensive survey
of linear algebra; it is meant only as a brief introduction and re-
0.6 Linear Trans-
view. The ideas and presentation order are modeled after Strang’s
forms
highly recommended Linear Algebra and its Applications.
At the heart of linear algebra is machinery for solving linear equa-
y
tions. In the simplest case, the number of unknowns equals the
number of equations. For example, here are a two equations in
2x−y=1
two unknowns:
(x,y)=(2,3)
2x − y = 1
x + y = 5.
(1)
x
x+y=5
There are at least two ways in which we can think of solving these
Figure 0.1 “Row” solu-
equations for x and y. The first is to consider each equation as
tion
describing a line, with the solution being at the intersection of the
lines: in this case the point (2, 3), Figure 0.1. This solution is
(1,5)
termed a “row” solution because the equations are considered in
isolation of one another. This is in contrast to a “column” solution
(−3,3)
in which the equations are rewritten in vector form:
(4,2)
2
1
x +
−1 y =
.
(2)
1
1
5
(−1,1)
(2,1)
The solution reduces to finding values for x and y that scale the
vectors (2, 1) and (
Figure 0.2 “Column”
−1,1) so that their sum is equal to the vector
(1, 5), Figure 0.2. Of course the solution is again x = 2 and y = 3.
solution
These solutions generalize to higher dimensions. Here is an exam-
ple with n = 3 unknowns and equations:
2u + v + w = 5
4u − 6v + 0w = −2
(3)
−2u + 7v + 2w = 9.
3
Each equation now corresponds to a plane, and the row solution
corresponds to the intersection of the planes (i.e., the intersection
of two planes is a line, and that line intersects the third plane at
a point: in this case, the point u = 1, v = 1, w = 2). In vector
form, the equations take the form:
2
1
1
5
4
−6
0
−2
−2u + 7 v + 2w = 9 .
(4)
The solution again amounts to finding values for u, v, and w that
scale the vectors on the left so that their sum is equal to the vector
on the right, Figure 0.3.
(5,−2,9)
Figure 0.3 “Column”
In the context of solving linear equations we have introduced the
solution
notion of a vector, scalar multiplication of a vector, and vector
sum. In its most general form, a n-dimensional column vector is
represented as:
x1x
2
x = . ,
(5)
.
.
x
n
and a n-dimensional row vector as:
y = ( y1 y2 . . . yn ) .
(6)
Scalar multiplication of a vector x by a scalar value c, scales the
length of the vector by an amount c (Figure 0.2) and is given by:
cv1.
cv = .. .
(7)
cv
n
The vector sum w = x + y is computed via the parallelogram
construction or by “stacking” the vectors head to tail (Figure 0.2)
and is computed by a pairwise addition of the individual vector
components:
w1 x1+y1
w2
x2 + y2
. =
.
.
(8)
.. .
.
w
n
xn+yn
The linear combination of vectors by vector addition and scalar
multiplication is one of the central ideas in linear algebra (more
on this later).
4
0.2 Matrices
In solving n linear equations in n unknowns there are three quan-
tities to consider. For example consider again the following set of
equations:
2u + v + w = 5
4u − 6v + 0w = −2
(9)
−2u + 7v + 2w = 9.
On the right is the column vector:
5
−29,
(10)
and on the left are the three unknowns that can also be written
as a column vector:
u
vw.
(11)
The set of nine coefficients (3 rows, 3 columns) can be written in
matrix form:
2 1 1
4 −6 0
−2 7 2
(12)
Matrices, like vectors, can be added and scalar multiplied. Not
surprising, since we may think of a vector as a skinny matrix: a
matrix with only one column. Consider the following 3×3 matrix:
a1 a2 a3
A =
a4 a5 a6
a7 a8 a9 .
(13)
The matrix cA, where c is a scalar value, is given by:
ca1 ca2 ca3
cA =
ca4 ca5 ca6
ca7 ca8 ca9 .
(14)
And the sum of two matrices, A = B + C, is given by:
a1 a2 a3
b1 + c1 b2 + c2 b3 + c3
a4 a5 a6
b4 + c4 b5 + c5 b6 + c6
a7 a8 a9 = b7 + c7 b8 + c8 b9 + c9 . (15)
5
With the vector and matrix notation we can rewrite the three
equations in the more compact form of Ax = b:
2 1 1u
5
4 −6 0 v
−2
−2 7 2w = 9 .
(16)
Where the multiplication of the matrix A with vector x must be
such that the three original equations are reproduced. The first
component of the product comes from “multiplying” the first row
of A (a row vector) with the column vector x as follows:
u
( 2 1
1 )
vw = (2u+1v+1w).
(17)
This quantity is equal to 5, the first component of b, and is simply
the first of the three original equations. The full product is com-
puted by multiplying each row of the matrix A with the vector x
as follows:
2 1 1u 2u+1v+1w 5
4 −6 0 v
4u − 6v + 0w
−2
−2 7 2w = −2u + 7v + 2w = 9 . (18)
In its most general form the product of a m × n matrix with a
n dimensional column vector is a m dimensional column vector
whose ith component is:
n
aijxj,
(19)
j=1
where aij is the matrix component in the ith row and jth column.
The sum along the ith row of the matrix is referred to as the inner
product or dot product between the matrix row (itself a vector) and
the column vector x. Inner products are another central idea in
linear algebra (more on this later). The computation for multi-
plying two matrices extends naturally from that of multiplying a
matrix and a vector. Consider for example the following 3 ×4 and
4 × 2 matrices:
b
a
11
b12
11
a12 a13 a14
b
A =
21
b22
a21 a22 a23 a24
.
(20)
b
a
31 b32
31
a32 a33 a34 and B =
b41 b42
The product C = AB is a 3 × 2 matrix given by:
a11b11 + a12b21 + a13b31 + a14b41
a11b12 + a12b22 + a13b32 + a14b42
a21b11 + a22b21 + a23b31 + a24b41
a21b12 + a22b22 + a23b32 + a24b42
.(21)
a31b11 + a32b21 + a33b31 + a34b41
a31b12 + a32b22 + a33b32 + a34b42
6
That is, the i, j component of the product C is computed from
an inner product of the ith row of matrix A and the jth column
of matrix B. Notice that this definition is completely consistent
with the product of a matrix and vector. In order to multiply
two matrices A and B (or a matrix and a vector), the column
dimension of A must equal the row dimension of B. In other words
if A is of size m × n, then B must be of size n × p (the product is
of size m × p). This constraint immediately suggests that matrix
multiplication is not commutative: usually AB = BA. However
matrix multiplication is both associative (AB)C = A(BC) and
distributive A(B + C) = AB + AC.
The identity matrix I is a special matrix with 1 on the diagonal
and zero elsewhere:
1 0 ... 0 0
0 1
. . .
0 0
I
= . .
.
.
(22)
..
. .
.
.
0 0
. . .
0 1
Given the definition of matrix multiplication, it is easily seen that
for any vector x, Ix = x, and for any suitably sized matrix, IA = A
and BI = B.
In the context of solving linear equations we have introduced the
notion of a vector and a matrix. The result is a compact notation
for representing linear equations, Ax = b. Multiplying both sides
by the matrix inverse A−1 yields the desired solution to the linear
equations:
A−1Ax = A−1b
Ix = A−1b
x = A−1b
(23)
A matrix A is invertible if there exists 1 a matrix B such that
BA = I and AB = I, where I is the identity matrix. The ma-
trix B is the inverse of A and is denoted as A−1. Note that this
commutative property limits the discussion of matrix inverses to
square matrices.
Not all matrices have inverses. Let’s consider some simple exam-
ples. The inverse of a 1 × 1 matrix A = ( a) is A−1 = ( 1/a );
but the inverse does not exist when a = 0. The inverse of a 2 × 2
1The inverse of a matrix is unique: assume that B and C are both the
inverse of matrix A, then by definition B = B(AC) = (BA)C = C, so that B
must equal C.
7
matrix can be calculated as:
a
b −1
1
d
=
−b ,
(24)
c
d
ad − bc −c a
but does not exist when ad − bc = 0. Any diagonal matrix is
invertible:
a1
1/a1
A =
.
.
..
and A−1 =
. .
, (25)
a
n
1/an
as long as all the diagonal components are non-zero. The inverse
of a product of matrices AB is (AB)−1 = B−1A−1. This is easily
proved using the associativity of matrix multiplication. 2 The
inverse of an arbitrary matrix, if it exists, can itself be calculated
by solving a collection of linear equations. Consider for example a
3 ×3 matrix A whose inverse we know must satisfy the constraint
that AA−1 = I:
2 1 1
1 0 0
4 −6 0
0 1 0
−2 7 2x1 x2 x3 = e1 e2 e3 = 0 0 1.(26)
This matrix equation can be considered “a column at a time”
yielding a system of three equations Ax1 = e1, Ax2 = e2, and
Ax3 = e3. These can be solved independently for the columns
of the inverse matrix, or simultaneously using the Gauss-Jordan
method.
A system of linear equations Ax = b can be solved by simply
left multiplying with the matrix inverse A−1 (if it exists). We
must naturally wonder the fate of our solution if the matrix is not
invertible. The answer to this question is explored in the next
section. But before moving forward we need one last definition.
The transpose of a matrix A, denoted as At, is constructed by
placing the ith row of A into the ith column of At. For example:
1
4
1
2
1
A =
and At = 2
4 −6 0
−6
1
0
(27)
In general, the transpose of a m ×n matrix is a n×m matrix with
(At)ij = Aji. The transpose of a sum of two matrices is the sum of
2In order to prove (AB)−1 = B−1A−1, we must show (AB)(B−1A−1) =
I: (AB)(B−1A−1) = A(BB−1)A−1 = AIA−1 = AA−1 = I, and that
(B−1A−1)(AB) = I: (B−1A−1)(AB) = B−1(A−1A)B = B−1IB = B−1B =
I.
8
the transposes: (A + B)t = At + Bt. The transpose of a product of
two matrices has the familiar form (AB)t = BtAt. And the trans-
pose of the inverse is the inverse of the transpose: (A−1)t = (At)−1.
Of particular interest will be the class of symmetric matrices that
are equal to their own transpose At = A. Symmetric matrices are
necessarily square, here is a 3 × 3 symmetric matrix:
2
1
4
A =
1 −6 0
4
0
3 ,
(28)
notice that, by definition, aij = aji.
0.3 Vector Spaces
The most common vector space is that defined over the reals, de-
noted as Rn. This space consists of all column vectors with n
real-valued components, with rules for vector addition and scalar
multiplication. A vector space has the property that the addi-
tion and multiplication of vectors always produces vectors that lie
within the vector space. In addition, a vector space must satisfy
the following properties, for any vectors x, y, z, and scalar c:
1. x + y = y + x
2. (x + y) + z = x + (y + z)
3. there exists a unique “zero” vector 0 such that x + 0 = x
4. there exists a unique “inverse” vector −x such that
x + (−x) = 0
5. 1x = x
6. (c1c2)x = c1(c2x)
7. c(x + y) = cx + cy
8. (c1 + c2)x = c1x + c2x
Vector spaces need not be finite dimensional, R∞ is a vector space.
Matrices can also make up a vector space. For example the space
of 3 × 3 matrices can be thought of as R9 (imagine stringing out
the nine components of the matrix into a column vector).
A subspace of a vector space is a non-empty subset of vectors that
is closed under vector addition and scalar multiplication. That
is, the following constraints are satisfied: (1) the sum of any two
vectors in the subspace remains in the subspace; (2) multiplication
of any vector by a scalar yields a vector in the subspace. With
the closure property verified, the eight properties of a vector space
automatically hold for the subspace.
Example 0.1
Consider the set of all vectors in R2 whose com-
ponents are greater than or equal to zero. The sum of any two
9
vectors in this space remains in the space, but multiplication of,
1
−1
for example, the vector
by −1 yields the vector
2
−2
which is no longer in the space. Therefore, this collection of
vectors does not form a vector space.
Vector subspaces play a critical role in understanding systems of
linear equations of the form Ax = b. Consider for example the
following system:
u1 v1
b
x
1
1
u2 v2
= b
x
2
u
2
3
v3
b3
(29)
Unlike the earlier system of equations, this system is over-constrained,
there are more equations (three) than unknowns (two). A solu-
tion to this system exists if the vector b lies in the subspace of the
columns of matrix A. To see why this is so, we rewrite the above
system according to the rules of matrix multiplication yielding an
equivalent form:
u1
v1
b1
x
1 u2
v2
b2
u3 + x2 v3 = b3 .
(30)
In this form, we see that a solution exists when the scaled columns
of the matrix sum to the vector b. This is simply the closure
property necessary for a vector subspace.
The vector subspace spanned by the columns of the matrix A is
called the column space of A. It is said that a solution to Ax = b
exists if and only if the vector b lies in the column space of A.
Example 0.2
Consider the following over-constrained system:
Ax
=
b
1
0
b
u
1
5
4
=
b
v
2
2
4
b3
The column space of A is the plane spanned by the vectors
( 1
5
2 )t and ( 0
4
4 )t. Therefore, the solution b can not
be an arbitrary vector in R3, but is constrained to lie in the
plane spanned by these two vectors.
At this point we have seen three seemingly different classes of
linear equations of the form Ax = b, where the matrix A is either:
1. square and invertible (non-singular),
10
2. square but not invertible (singular),
3. over-constrained.
In each case solutions to the system exist if the vector b lies in the
column space of the matrix A. At one extreme is the invertible
n×n square matrix whose solutions may be any vector in the whole
of Rn. At the other extreme is the zero matrix A = 0 with only
the zero vector in it’s column space, and hence the only possible
solution. In between are the singular and over-constrained cases,
where solutions lie in a subspace of the full vector space.
The second important vector space is the nullspace of a matrix.
The vectors that lie in the nullspace of a matrix consist of all
solutions to the system Ax = 0. The zero vector is always in the
nullspace.
Example 0.3
Consider the following system:
Ax
=
0
1
0
1
u
0
5
4
9
v
=
0
2
4
6
w
0
The nullspace of A contains the zero vector ( u
v
w )t = ( 0
0
0 )t.
Notice also that the third column of A is the sum of the first two
columns, therefore the nullspace of A also contains all vectors of
the form ( u
v
w )t = ( c
c
−c )t (i.e., all vectors lying on a
one-dimensional line in R3).
0.4 Basis
(2,2)
Recall that if the matrix A in the system Ax = b is invertible, then
left multiplying with A−1 yields the desired solution: x = A−1b.
(−1,−1)
In general it is said that a n × n matrix is invertible if it has rank
n or is full rank, where the rank of a matrix is the number of
linearly independent rows in the matrix. Formally, a set of vectors
(2,2)
u1, u2, ..., un are linearly independent if:
c1u1 + c2u2 + ... + cnun = 0
(31)
(−2,0)
is true only when c1 = c2 = ... = cn = 0. Otherwise, the vectors
are linearly dependent. In other words, a set of vectors are linearly
(−1,2)
dependent if at least one of the vectors can be expressed as a sum
(2,2)
of scaled copies of the remaining vectors.
(−2,0)
Linear independence is easy to visualize in lower-dimensional sub-
spaces. In 2-D, two vectors are linearly dependent if they lie along
a line, Figure 0.4. That is, there is a non-trivial combination of the
Figure 0.4 Linearly de-
pendent
11
(top/bottom) and inde-
pendent (middle).
vectors that yields the zero vector. In 2-D, any three vectors are
guaranteed to be linearly dependent. For example, in Figure 0.4,
the vector ( −1 2) can be expressed as a sum of the remaining
linearly independent vectors: 3 (
2
−2 0) + (2 2). In 3-D, three
vectors are linearly dependent if they lie in the same plane. Also
in 3-D, any four vectors are guaranteed to be linearly dependent.
Linear independence is directly related to the nullspace of a ma-
trix. Specifically, the columns of a matrix are linearly independent
(i.e., the matrix is full rank) if the matrix nullspace contains only
the zero vector. For example, consider the following system of
linear equations:
u1 v1 w1x1
0
u2 v2 w2 x2
0
u3 v3 w3 x3 = 0.
(32)
Recall that the nullspace contains all vectors x such that x1u +
x2v + x3w = 0. Notice that this is also the condition for linear
independence. If the only solution is the zero vector then the
vectors are linearly independent and the matrix is full rank and
invertible.
Linear independence is also related to the column space of a ma-
trix. If the column space of a n × n matrix is all of Rn, then the
matrix is full rank. For example, consider the following system of
linear equations:
u1 v1 w1x1
b1
u2 v2 w2 x2
b2
u3 v3 w3 x3 = b3 .
(33)
If the the matrix is full rank, then the solution b can be any vector
in R3. In such cases, the vectors u, v, w are said to span the space.
Now, a linear basis of a vector space is a set of linearly independent
vectors that span the space. Both conditions are important. Given
an n dimensional vector space with n basis vectors v1, ..., vn, any
vector u in the space can be written as a linear combination of
these n vectors:
u = a1v1 + ... + anvn.
(34)
In addition, the linear independence guarantees that this linear
combination is unique. If there is another combination such that:
u = b1v1 + ... + bnvn,
(35)
12
then the difference of these two representations yields
0 = (a1 − b1)v1 + ... + (an − bn)vn
= c1v1 + ... + cnvn
(36)
which would violate the linear independence condition. While
the representation is unique, the basis is not. A vector space has
infinitely many different bases. For example in R2 any two vectors
that do not lie on a line form a basis, and in R3 any three vectors
that do not lie in a common plane or line form a basis.
Example 0.4
The vectors ( 1
0 ) and ( 0
1 ) form the canonical
basis for R2. These vectors are both linearly independent and
span the entire vector space.
Example 0.5
The vectors ( 1
0
0 ), ( 0
1
0 ) and ( −1
0
0 )
do not form a basis for R3. These vectors lie in a 2-D plane and
do not span the entire vector space.
Example 0.6
The vectors ( 1
0
0 ), ( 0
−1
0 ), ( 0
0
2 ),
and ( 1
−1
0 ) do not form a basis. Although these vectors
span the vector space, the fourth vector is linearly dependent on
the first two. Removing the fourth vector leaves a basis for R3.
0.5 Inner Products and Projections
0.6 Linear Transforms
13
1. Discrete-Time Signals and Systems
1.1 Discrete-Time
1.1 Discrete-Time Signals
Signals
A discrete-time signal is represented as a sequence of numbers, f ,
1.2 Discrete-Time
where the xth number in the sequence is denoted as f [x]:
Systems
f
= {f[x]},
−∞ < x < ∞,
(1.1)
1.3 Linear
Time-
where x is an integer. Note that from this definition, a discrete-
Invariant Sys-
time signal is defined only for integer values of x. For example,
tems
the finite-length sequence shown in Figure 1.1 is represented by
the following sequence of numbers
f[x]
f
= {f[1] f[2] ... f[12] }
=
x
{0 1 2 4 8 7 6 5 4 3 2 1}. (1.2)
Figure 1.1
For notational convenience, we will often drop the cumbersome
Discrete-time signal
notation of Equation (1.1), and refer to the entire sequence sim-
ply as f [x]. Discrete-time signals often arise from the periodic
sampling of continuous-time (analog) signals, a process that we
will cover fully in later chapters.
1.2 Discrete-Time Systems
In its most general form, a discrete-time system is a transformation
that maps a discrete-time signal, f [x], onto a unique g[x], and is
f[x]
T
g[x]
denoted as:
Figure 1.2
g[x] = T {f[x]}
(1.3)
Discrete-time system
Example 1.1
Consider the following system:
N
1
g[x]
=
f [x + k].
2N + 1 k=−N
In this system, the kth number in the output sequence is com-
puted as the average of 2N + 1 elements centered around the kth
f[x]
input element. As shown in Figure 1.3, with N = 2, the output
value at k = 5 is computed as 1/5 times the sum of the five input
x
elements between the dotted lines. Subsequent output values are
3
5
7
computed by sliding these lines to the right.
Figure 1.3 Moving
Av-
erage
Although in the above example, the output at each position k
depended on only a small number of input values, in general, this
may not be the case, and the output may be a function of all input
values.
14
1.3 Linear Time-Invariant Systems
Of particular interest to us will be a class of discrete-time systems
that are both linear and time-invariant. A system is said to be
linear if it obeys the rules of superposition, namely:
T {af1[x] + bf2[x]} = aT{f1[x]} + bT{f2[x]},
(1.4)
for any constants a and b. A system, T {·} that maps f[x] onto g[x]
is shift- or time-invariant if a shift in the input causes a similar
shift in the output:
g[x] = T {f[x]}
=⇒
g[x − x0] = T{f[x − x0]}. (1.5)
Example 1.2
Consider the following system:
g[x]
=
f [x] − f [x − 1],
−∞ < x < ∞.
In this system, the kth number in the output sequence is com-
puted as the difference between the kth and kth-1 elements in
f[x]
the input sequence. Is this system linear? We need only show
that this system obeys the principle of superposition:
T {af1[x] + bf2[x]}
=
(af1[x] + bf2[x]) − (af1[x − 1] + bf2[x − 1])
x
=
(af1[x] − af1[x − 1]) + (bf2[x] − bf2[x − 1])
g[x]
=
a(f1[x] − f1[x − 1]) + b(f2[x] − f2[x − 1])
which, according to Equation (1.4), makes T {·} linear. Is this
x
system time-invariant? First, consider the shifted signal, f1[x] =
f [x − x0], then:
Figure 1.4 Backward
difference
g1[x] = f1[x] − f1[x − 1] = f [x − x0] − f [x − 1 − x0],
and,
g[x − x0] = f [x − x0] − f [x − 1 − x0] = g1[x],
so that this system is time-invariant.
Example 1.3
Consider the following system:
g[x]
=
f [nx],
−∞ < x < ∞,
where n is a positive integer. This system creates an output
sequence by selecting every nth element of the input sequence.
Is this system linear?
T {af1[x] + bf2[x]}
=
af1[nx] + bf2[nx]
which, according to Equation (1.4), makes T {·} linear. Is this
system time-invariant? First, consider the shifted signal, f1[x] =
f [x − x0], then:
g1[x] = f1[nx] = f [nx − x0],
but,
g[x − x0] = f [n(x − x0)] = g1[x],
so that this system is not time-invariant.
15
The precise reason why we are particularly interested in linear
time-invariant systems will become clear in subsequent chapters.
But before pressing on, the concept of discrete-time systems is
reformulated within a linear-algebraic framework. In order to ac-
complish this, it is necessary to first restrict ourselves to consider
input signals of finite length. Then, any discrete-time linear sys-
tem can be represented as a matrix operation of the form:
g = M f ,
(1.6)
where, f is the input signal, g is the output signal, and the matrix
M embodies the discrete-time linear system.
Example 1.4
Consider the following system:
g[x]
=
f [x − 1],
1 < x < 8.
The output of this system is a shifted copy of the input signal,
and can be formulated in matrix notation as:
g[1]
0
0
0
0
0
0
0
1
f [1]
g[2]
1 0 0 0 0 0 0 0f[2]
g[3]
0
1
0
0
0
0
0
0
f [3]
g[4] 0 0 1 0 0 0 0 0f[4]
=
g[5] 0 0 0 1 0 0 0 0f[5]
g[6] 0 0 0 0 1 0 0 0f[6]
g[7] 0 0 0 0 0 1 0 0f[7]
g[8]
0 0 0 0 0 0 1 0f[8]
Note that according to the initial definition of the system, the
output signal at x = 1 is undefined (i.e., g[1] = f [0]). In the
above matrix formulation we have adopted a common solution
to this problem by considering the signal as wrapping around
itself and setting g[1] = f [8].
Any system expressed in the matrix notation of Equation (1.6) is
a discrete-time linear system, but not necessarily a time-invariant
system. But, if we constrain ourselves to Toeplitz matrices, then
the resulting expression will be a linear time-invariant system. A
Toeplitz matrix is one in which each row contains a shifted copy
of the previous row. For example, a 5 ×5 Toeplitz matrix is of the
form
m1 m2 m3 m4 m5
m
5
m1 m2 m3 m4
M
= m4 m5 m1 m2 m3
(1.7)
m
3 m4 m5 m1 m2
m
2
m3 m4 m5 m1
It is important to feel comfortable with this formulation because
later concepts will be built upon this linear algebraic framework.
16
2. Linear Time-Invariant Systems
Our interest in the class of linear time-invariant systems (LTI) is
2.1 Space: Convo-
motivated by the fact that these systems have a particularly con-
lution Sum
venient and elegant representation, and this representation leads
us to several fundamental tools in signal and image processing.
2.2 Frequency:
Fourier Trans-
2.1 Space: Convolution Sum
form
In the previous section, a discrete-time signal was represented as
a sequence of numbers. More formally, this representation is in
terms of the discrete-time unit impulse defined as:
1
x
1, x = 0
δ[x] =
(2.1)
0, x = 0.
Figure 2.1 Unit impulse
Any discrete-time signal can be represented as a sum of scaled and
shifted unit-impulses:
∞
f [x] =
f [k]δ[x − k],
(2.2)
k=−∞
where the shifted impulse δ[x − k] = 1 when x = k, and is zero
elsewhere.
Example 2.1
Consider the following discrete-time signal, cen-
tered at x = 0.
f [x]
=
( . . .
0
0
2
−1
4
0
0
. . . ) ,
this signal can be expressed as a sum of scaled and shifted unit-
impulses:
f [x]
=
2δ[x + 1] − 1δ[x] + 4δ[x − 1]
=
f [−1]δ[x + 1] + f [0]δ[x] + f [1]δ[x − 1]
1
=
f [k]δ[x − k].
k=−1
Let’s now consider what happens when we present a linear time-
invariant system with this new representation of a discrete-time
signal:
g[x] = T {f[x]}
∞
= T
f[k]δ[x−k].
(2.3)
k=−∞
17
By the property of linearity, Equation (1.4), the above expression
may be rewritten as:
∞
g[x] =
f [k]T {δ[x − k]}.
(2.4)
k=−∞
Imposing the property of time-invariance, Equation (1.5), if h[x] is
the response to the unit-impulse, δ[x], then the response to δ[x−k]
is simply h[x−k]. And now, the above expression can be rewritten
as:
∞
g[x] =
f [k]h[x − k].
(2.5)
k=−∞
Consider for a moment the implications of the above equation.
The unit-impulse response, h[x] = T {δ[x]}, of a linear time-invariant
system fully characterizes that system. More precisely, given the
unit-impulse response, h[x], the output, g[x], can be determined
for any input, f [x].
The sum in Equation (2.5) is commonly called the convolution
sum and may be expressed more compactly as:
g[x] = f [x] ⋆ h[x].
(2.6)
A more mathematically correct notation is (f ⋆ h)[x], but for later
notational considerations, we will adopt the above notation.
Example 2.2
Consider the following finite-length unit-impulse
response:
h[x]
=
( −2
4
−2 ) ,
h[x]
0
and the input signal, f [x], shown in Figure 2.2. Then the output
signal at, for example, x = −2, is computed as:
−1
g[−2]
=
f [k]h[−2 − k]
f[x]
0
k=−3
=
f [−3]h[1] + f [−2]h[0] + f [−1]h[−1].
g[−2]
Figure 2.2 Convolution:
The next output sum at x = −1, is computed by “sliding” the
g[x] = f [x] ⋆ h[x]
unit-impulse response along the input signal and computing a
similar sum.
Since linear time-invariant systems are fully characterized by con-
volution with the unit-impulse response, properties of such sys-
tems can be determined by considering properties of the convolu-
tion operator. For example, convolution is commutative:
∞
f [x] ⋆ h[x] =
f [k]h[x − k],
let j = x − k
k=−∞
18
∞
∞
=
f [x − j]h[j] =
h[j]f [x − j]
j=−∞
j=−∞
= h[x] ⋆ f [x].
(2.7)
Convolution also distributes over addition:
∞
f [x] ⋆ (h1[x] + h2[x]) =
f [k](h1[x − k] + h2[x − k])
k=−∞
∞
=
f [k]h1[x − k] + f[k]h2[x − k]
k=−∞
∞
∞
=
f [k]h1[x − k] +
f [k]h2[x − k]
k=−∞
k=−∞
= f [x] ⋆ h1[x] + f [x] ⋆ h2[x].
(2.8)
A final useful property of linear time-invariant systems is that
a cascade of systems can be combined into a single system with
f[x]
h1[x]
h2[x]
g[x]
impulse response equal to the convolution of the individual impulse
responses. For example, for a cascade of two systems:
f[x]
h1[x] * h2[x]
g[x]
(f [x] ⋆ h1[x]) ⋆ h2[x] = f [x] ⋆ (h1[x] ⋆ h2[x]).
(2.9)
Figure 2.3 Identical
This property is fairly straight-forward to prove, and offers a good
LTIs
exercise in manipulating the convolution sum:
g1[x] = f [x] ⋆ h1[x]
∞
=
f [k]h1[x − k] and,
k=−∞
(2.10)
g2[x] = (f [x] ⋆ h1[x]) ⋆ h2[x]
= g1[x] ⋆ h2[x]
∞
=
g1[j]h2[x − j] substituting for g1[x],
j=−∞
∞
∞
=
f[k]h1[j −k]
j=−∞ k=−∞
h2[x−j]
∞
∞
=
f [k]
h1[j − k]h2[x − j]
k=−∞
j=−∞
let i= j −k,
∞
∞
=
f [k]
h1[i]h2[x − i − k]
k=−∞
i=−∞
= f [x] ⋆ (h1[x] ⋆ h2[x]).
(2.11)
Let’s consider now how these concepts fit into the linear-algebraic
framework. First, a length N signal can be thought of as a point
19
in a N −dimensional vector space. As a simple example, consider
the length 2 signal shown in Figure 2.4, represented as a vector
in a 2-dimensional space. Earlier, the signal f [x] was expressed
as a sum of weighted and shifted impulses, f [x] = 9δ[x] + 4δ[x −
f[x]
1], and in the vector space, it is expressed with respect to the
9
canonical basis as f = 9 ( 1 0 ) + 4 ( 0 1 ). The weighting of each
4
basis vector is determined by simply projecting the vector f onto
each axis. With this vector representation, it is then natural to
x
express the convolution sum (i.e., a linear time-invariant system)
as a matrix operation. For example, let h[x] = ( h−1 h0 h1 ) be
(0,1)
the finite-length unit-impulse response of a linear time-invariant
system, T {·}, then the system g[x] = T{f[x]} can be expressed as
f = (9,4)
g = M f , where the matrix M is of the form:
(0,4)
h0
h 1
0
0
. . .
0
0
0
h1
−
(1,0)
h1 h0 h 1 0 ... 0 0 0 0
(9,0)
−
0
h1
h0
h 1
. . .
0
0
0
0
−
Figure 2.4
M
=
..
.
.
.
. .
.
.
,
(2.12)
Signal and vector repre-
0
0
0
0
. . . h1
h0
h 1
0
sentation
−
0 0 0 0 ... 0 h1 h0 h 1
−
h 1
0
0
0
. . .
0
0
h1
h0
−
where each row contains a shifted and time-reversed copy of the
unit-impulse response, h[x]. The convolution matrix can then be
thought of as simply transforming the basis set. As expected,
this matrix is a Toeplitz matrix of the form given earlier in Equa-
tion (1.7). The reason for the time-reversal can be seen directly
from the convolution sum of Equation (2.5). More specifically,
the output signal g[x] at a fixed x, is determined by summing the
products of f [k]h[x −k] for all k. Note that the signal h[x−k] can
be equivalently written as h[−k +x], which is a shifted (by x) and
time-reversed (because of the −k) copy of the impulse response.
Note also that when expressed in matrix form, it becomes imme-
diately clear that the convolution sum is invertible, when h is not
identically zero: g = M f and f = M −1g.
Before pressing on, let’s try to combine the main ideas seen so far
f[x]
into a single example. We will begin by defining a simple discrete-
time system, show that it is both linear and time-invariant, and
compute its unit-impulse response
g[x]
Example 2.3
Define the discrete-time system, T {·} as:
f[x]
g[x]
=
f [x + 1] − f [x − 1].
This system is linear because it obeys the rule of superposition:
T {af1[x] + bf2[x]}
=
(af1[x + 1] + bf2[x + 1]) − (af1[x − 1] + bf2[x − 1])
=
(af1[x + 1] − af1[x − 1]) + (bf2[x + 1] − bf2[x − 1])
g[x]
=
a(f1[x + 1] − f1[x − 1]) + b(f2[x + 1] − f2[x − 1])
Figure 2.5 g[x] = f [x] ⋆
h[x]
20
This system is also time-invariant because a shift in the input,
f1[x] = f [x − x0], leads to a shift in the output:
g1[x]
=
f1[x + 1] − f1[x − 1]
=
f [x + 1 − x0] − f [x − 1 − x0]
and,
g[x − x0]
=
f [x + 1 − x0] − f [x − 1 − x0]
=
g1[x].
The unit-impulse response is given by:
h[x]
=
T {δ[x]}
=
δ[x + 1] − δ[x − 1]
=
( . . .
0
1
0
−1
0
. . . ) .
So, convolving the finite-length impulse response h[x] = ( 1
0
−1 )
with any input signal, f [x], gives the output of the linear time-
invariant system, g[x] = T {f [x]}:
∞
x+1
g[x]
=
f [k]h[x − k] =
f [k]h[x − k].
k=−∞
k=x−1
And, in matrix form, this linear time-invariant system is given
by g = M f , where:
0 1 0 0 ... 0 0 0 −1
−1
0
1
0
. . .
0
0
0
0
0 −1 0 1 ... 0 0 0 0
.
.
M
=
.
.
. .
. .
.
.
0 0 0 0 ... −1 0 1 0
0 0 0 0 ... 0 −1 0 1
1
0
0
0
. . .
0
0
−1
0
2.2 Frequency: Fourier Transform
1
In the previous section the expression of a discrete-time signal as
a sum of scaled and shifted unit-impulses led us to the convolution
0
sum. In a similar manner, we will see shortly how expressing a
signal in terms of sinusoids will lead us to the Fourier transform,
−1
and then to a new way to think about discrete-time systems. The
1
basic signal used as the building block is the sinusoid:
0
A cos[ωx + φ],
−∞ < x < ∞,
(2.13)
where A is the amplitude, ω is the frequency, and φ is the phase of
−1
the sinusoid. Shown in Figure 2.6, from top to bottom, are cos[x],
1
cos[2x], and cos[x + π/2]. Consider next the following, seemingly
unrelated, complex exponential eiωx with frequency ω, and i the
0
complex value √−1. This function is simply a compact notation
for describing a sum of the sine and cosine function:
−1
Aeiωx = A cos(ωx) + iA sin(ωx).
(2.14)
Figure 2.6
f [x] = A cos[ωx + φ]
21
The complex exponential has a special relationship with linear
time-invariant systems - the output of a linear time-invariant sys-
tem with unit-impulse response h[x] and a complex exponential as
input is:
g[x] = eiωx ⋆ h[x]
∞
=
h[k]eiω(x−k)
k=−∞
∞
= eiωx
h[k]e−iωk
(2.15)
k=−∞
Defining H[ω] to be the summation component, g[x] can be ex-
pressed as:
g[x] = H[ω]eiωx,
(2.16)
that is, given a complex exponential as input, the output of a
linear time-invariant system is again a complex exponential of the
Imaginary
same frequency scaled by some amount. 3 The scaling of the com-
plex exponential, H[w], is called the frequency response and is
H = R + I
generally a complex-valued function expressed in terms of its real
| H |
and imaginary components:
H[ω] = HR[ω] + iHI[ω],
(2.17)
H
Real
or more commonly in terms of the magnitude and phase:
Figure 2.7 Magnitude
and phase
|H[ω]| = HR[ω]2 + HI[ω]2 and ≺ H[ω] = tan−1 HI[(ω] .
HR[ω]
Example 2.4
Consider the following linear time-invariant sys-
tem, T {·}:
g[x]
=
f [x − x0].
This system outputs a time-delayed copy of the input signal.
The frequency response of this system can be determined by
considering the input signal f [x] = eiωx:
g[x]
=
eiω(x−x0)
=
e−iωx0 eiωx,
which is of the same form as Equation (2.16), with frequency
response H[ω] = e−iωx0 . Decomposing this response in terms of
the real and imaginary components gives:
HR[ω] = cos[ωx0]
and
HI [ω] = − sin[ωx0],
3In linear algebraic terms, the complex exponentials are said to be the
eigenfunctions of LTIs, and H[ω] the eigenvalue.
22
or in terms of the magnitude and phase:
|H[ω]|
=
cos2[ωx0] + − sin2[ωx0]
=
1
≺ H[ω]
=
tan−1
− sin[ωx0]
cos[ωx0]
=
−ωx0.
Intuitively, this should make perfect sense. This system simply
takes an input signal and outputs a delayed copy, therefore, there
is no change in the magnitude of each sinusoid, while there is a
phase shift proportional to the delay, x0.
So, why the interest in sinusoids and complex exponentials? As
we will show next, a broad class of signals can be expressed as a
linear combination of complex exponentials, and analogous to the
impulse response, the frequency response completely characterizes
the system.
Let’s begin by taking a step back to the more familiar sinusoids,
and then work our way to the complex exponentials. Any periodic
discrete-time signal, f [x], can be expressed as a sum of scaled,
phase-shifted sinusoids of varying frequencies:
1
π
f [x] =
c
2π
k cos [kx + φk]
− ∞ < x < ∞, (2.18)
k=−π
For each frequency, k, the amplitude is a real number, ck ∈ R,
and the phase, φk ∈ [0,2π]. This expression is commonly referred
to as the Fourier series.
Example 2.5
Shown below is a signal, f [x] (left) represented as
a sum of the first four sinusoids: f [x] = c0 cos[0x + φ0] + ... +
c3 cos[3x + φ3].
=
c0
+ c1
+
c2
+ c3
.
In the language of linear algebra, the sinusoids are said to form
a basis for the set of periodic signals, that is, any periodic signal
can be written as a linear combination of the sinusoids. Recall
23
that in deriving the convolution sum, the basis consisted of shifted
copies of the unit-impulse. But note now that this new basis is
not fixed because of the phase term, φk. It is, however, possible
to rewrite the Fourier series with respect to a fixed basis of zero-
phase sinusoids. With the trigonometric identity cos(A + B) =
cos(A) cos(B)−sin(A)sin(B), the Fourier series of Equation (2.18)
may be rewritten as:
1
π
f [x] =
c
2π
k cos[kx + φk]
k=−π
1
π
=
c
2π
k cos[φk] cos[kx] + ck sin[φk] sin[kx]
k=−π
1
π
=
a
2π
k cos[kx] + bk sin[kx]
(2.19)
k=−π
In this expression, the constants ak and bk are the Fourier coef-
ficients and are determined by the Fourier transform. In other
words, the Fourier transform simply provides a means for express-
ing a signal in terms of the sinusoids. The Fourier coefficients are
given by:
∞
∞
ak =
f [j] cos[kj]
and
bk =
f [j] sin[kj]
(2.20)
j=−∞
j=−∞
Notice that the Fourier coefficients are determined by projecting
the signal onto each of the sinusoidal basis. That is, consider both
the signal f [x] and each of the sinusoids as T -dimensional vectors,
f and b, respectively. Then, the projection of f onto b is:
f0b0 + f1b1 + ... =
fjbj,
(2.21)
j
where the subscript denotes the jth entry of the vector.
Often, a more compact notation is used to represent the Fourier
series and Fourier transform which exploits the complex exponen-
tial and its relationship to the sinusoids:
eiωx = cos(ωx) + i sin(ωx),
(2.22)
where i is the complex value √−1. Under the complex exponential
notation, the Fourier series and transform take the form:
1
π
∞
f [x] =
c
f [j]e−ikj,
(2.23)
2π
keikx
and ck =
k=−π
j=−∞
24
where ck = ak − ibk. This notation simply bundles the sine and
cosine terms into a single expression. A more common, but equiv-
alent, notation is:
1
π
∞
f [x] =
F [ω]eiωx
and F [ω] =
f [k]e−iωk. (2.24)
2π ω=−π
k=−∞
Comparing the Fourier transform (Equation (2.24)) with the fre-
d[x]
exp(iwx)
quency response (Equation (2.16)) we see now that the frequency
response of a linear time-invariant system is simply the Fourier
transform of the unit-impulse response:
LTI
LTI
∞
Fourier
H[ω] =
h[k]e−iωk.
(2.25)
h[x]
H[w]exp(iwx)
k=−∞
Transform
In other words, linear time-invariant systems are completely char-
g[x]=f[x]*h[x]
G[w]=F[w]H[w]
acterized by their impulse response, h[x], and, equivalently, by
Figure 2.8 LTI:
space
their frequency response, H[ω], Figure 2.8.
and frequency
Example 2.6
Consider the following linear time-invariant sys-
tem, T {·}:
1
1
1
g[x]
=
f [x − 1] +
f [x] +
f [x + 1].
4
2
4
The output of this system at each x, is a weighted average of
the input signal centered at x. First, let’s compute the impulse
response:
1
1
1
h[x]
=
δ[x − 1] +
δ[x] +
δ[x + 1]
4
2
4
=
( . . .
0
0
1
1
1
0
0
. . . ) .
4
2
4
Then, the frequency response is the Fourier transform of this
impulse response:
∞
H[ω]
=
h[k]e−iωk
k=−∞
1
=
h[k]e−iωk
k=−1
1
1
1
=
eiω +
e0 +
e−iω
4
2
4
1
1
1
=
(cos(ω) + i sin(ω)) +
+
(cos(ω) − i sin(ω))
4
2
4
1
1
=
|
+
cos(ω) |.
2
2
In this example, the frequency response is strictly real (i.e., HI [ω] =
0) and the magnitude and phase are:
|H[ω]|
=
HR[ω]2 + HI [ω]2
1
1
=
+
cos(ω)
2
2
≺ H[ω]
=
tan−1
HI [ω]
Hr[ω]
=
0
25
Both the impulse response and the magnitude of the frequency
response are shown below, where for clarity, the frequency re-
sponse was drawn as a continuous function.
Space (h[x])
Frequency (|H[ω]|)
1
0.5
0.25
0
0
−1
0
1
−pi
0
pi
Example 2.7
Consider the following linear time-invariant sys-
tem, T {·}:
1
1
g[x]
=
f [x + 1] −
f [x − 1].
2
2
The output of this system at each x, is the difference between
neighboring input values. The impulse response of this system
is:
1
1
h[x]
=
δ[x + 1] −
δ[x − 1]
2
2
=
( . . .
0
0
1
0
− 1
0
0
. . . ) .
2
2
Then, the frequency response is the Fourier transform of this
impulse response:
∞
H[ω]
=
h[k]e−iωk
k=−∞
1
=
h[k]e−iωk
k=−1
1
1
=
eiω + 0e0 −
e−iω
2
2
1
1
=
(cos(ω) + i sin(ω)) −
(cos(ω) − i sin(ω))
2
2
=
i sin(ω).
In this example, the frequency response is strictly imaginary
(i.e., HR[ω] = 0) because the impulse response is anti-symmetric,
and the magnitude and phase are:
|H[ω]|
=
HR[ω]2 + HI[ω]2
=
| sin(ω) |
≺ H[ω]
=
tan−1
HI [ω]
Hr[ω]
π
=
.
2
Both the impulse response and the magnitude of the frequency
response are shown below, where for clarity, the frequency re-
sponse was drawn as a continuous function.
26
Space (h[x])
Frequency (|H[ω]|)
1
0.5
0
−0.5
0
−1
0
1
−pi
0
pi
This system is an (approximate) differentiator, and can be seen
from the definition of differentiation:
df (x)
f (x + ε) − f (x − ε)
=
lim
,
dx
ε→0
ε
where, in the case of the system T {·}, ε is given by the dis-
tance between samples of the discrete-time signal f [x]. Let’s
see now if we can better understand the frequency response of
this system, recall that the magnitude was given by | sin(ω)| and
the phase by π . Consider now the derivative of a fixed fre-
2
quency sinusoid sin(ωx), differentiating with respect to x gives
ω cos(ωx) = ω sin(ωx − π/2). Note that differentiation causes a
phase shift of π/2 and a scaling by the frequency of the sinusoid.
Notice that this is roughly in-line with the Fourier transform, the
difference being that the amplitude is given by | sin(ω)| instead
of ω. Note though that for small ω, | sin(ω)| ≈ ω. This dis-
crepancy makes the system only an approximate, not a perfect,
differentiator.
Linear time-invariant systems can be fully characterized by their
impulse, h[x], or frequency responses, H[ω], both of which may be
used to determine the output of the system to any input signal,
f [x]:
g[x] = f [x] ⋆ h[x]
and
G[ω] = F [ω]H[ω],
(2.26)
where the output signal g[x] can be determined from its Fourier
transform G[ω], by simply applying the inverse Fourier transform.
This equivalence illustrates an important relationship between the
space and frequency domains. Namely, convolution in the space
domain is equivalent to multiplication in the frequency domain.
This is fairly straight-forward to prove:
g[x] = f [x] ⋆ h[x]
Fourier transforming,
∞
∞
g[k]e−iωk =
(f [k] ⋆ h[k])e−iωk
k=−∞
k=−∞
∞
∞
G[ω] =
f[j]h[k−j]
k=−∞ j=−∞
e−iωk
27
∞
∞
=
f [j]
h[k − j]e−iωk
let l = k − j,
j=−∞
k=−∞
∞
∞
=
f [j]
h[l]e−iω(l+j)
j=−∞
l=−∞
∞
∞
=
f [j]e−iωj
h[l]e−iωl
j=−∞
l=−∞
= F [ω]H[ω].
(2.27)
Like the convolution sum, the Fourier transform can be formulated
as a matrix operation:
F[0]
1
1
1
. . .
1
f [0]
F [ω]
f[1]
e−0i
e−i
e−2i
. . .
e−T i
F[2ω] = e−0i e−2i e−4i ... e−2Ti f[2]
. .
.
.
.
. .
. .
.
.
. .
.
.
F [T ω]
e−0i e−Ti e−2Ti ... e−T2if[T]
F
= M f .
(2.28)
Notice that this equation embodies both the Fourier transform and
the Fourier series of Equation (2.24). The above form is the Fourier
transform, and the Fourier series is gotten by left-multiplying with
the inverse of the matrix, M −1F = f .
28
3. Sampling: Continuous to Discrete (and back)
It is often more convenient to process a continuous-time signal
3.1 C/D: Space
with a discrete-time system. Such a system may consist of three
distinct stages: (1) the conversion of a continuous-time signal to a
3.2 C/D:
discrete-time signal (C/D converter); (2) the processing through a
Frequency
discrete-time system; and (3) the conversion of the output discrete-
time signal back to a continuous-time signal (D/C converter). Ear-
3.3 D/C
lier we focused on the discrete-time processing, and now we will
concentrate on the conversions between discrete- and continuous-
f(x)
time signals. Of particular interest is the somewhat remarkable
fact that under certain conditions, a continuous-time signal can
be fully represented by a discrete-time signal!
C/D
3.1 Continuous to Discrete: Space
f[x]
A discrete-time signal, f [x], is formed from a continuous-time sig-
nal, f (x), by the following relationship:
T
f [x] = f (xT )
− ∞ < x < ∞,
(3.1)
g[x]
for integer values x. In this expression, the quantity T is the
sampling period. In general, continuous-time signals will be de-
D/C
noted with rounded parenthesis (e.g., f (·)), and discrete-time sig-
nals with square parenthesis (e.g., f [·]). This sampling operation
g(x)
may be considered as a multiplication of the continuous time sig-
Figure 3.1 Processing
nal with an impulse train, Figure 3.2. The impulse train is defined
block diagram
as:
∞
s(x) =
δ(x
f(x)
− kT),
(3.2)
k=−∞
where δ(
x
·) is the unit-impulse, and T is the sampling period - note
that the impulse train is a continuous-time signal. Multiplying
f[x]
the impulse train with a continuous-time signal gives a sampled
signal:
f
Figure 3.2 Sampling:
s(x)
= f (x)s(x),
(3.3)
space
Note that the sampled signal, fs(x), is indexed on the continuous
variable x, while the final discrete-time signal, f [x] is indexed on
the integer variable x. It will prove to be mathematically conve-
nient to work with this intermediate sampled signal, fs(x).
29
3.2 Continuous to Discrete: Frequency
In the space domain, sampling was described as a product between
F(w)
the impulse train and the continuous-time signal (Equation (3.3)).
In the frequency domain, this operation amounts to a convolution
w
between the Fourier transform of these two signals:
−w n
w n
S(w)
Fs(ω) = F (ω) ⋆ S(ω)
(3.4)
For example, shown in Figure 3.3 (from top to bottom) are the
w
−w
0
w
s
s
Fourier transforms of the continuous-time function, F (ω), the im-
Fs(w)
pulse train, S(ω), itself an impulse train, and the results of con-
volving these two signals, Fs(ω). Notice that the Fourier trans-
form of the sampled signal contains multiple (yet exact) copies
w
of the Fourier transform of the original continuous signal. Note
w s− w n
however the conditions under which an exact replica is preserved
Figure 3.3 Sampling:
depends on the maximum frequency response ωn of the original
no aliasing
continuous-time signal, and the sampling interval of the impulse
train, ωs which, not surprisingly, is related to the sampling pe-
riod T as ωs = 2π/T . More precisely, the copies of the frequency
response will not overlap if:
F(w)
ωn < ωs − ωn
or
ωs > 2ωn,
(3.5)
w
−w n
w n
The frequency ωn is called the Nyquist frequency and 2ωn is called
S(w)
the Nyquist rate.
Shown in Figure 3.4 is another example of
this sampling process in the frequency domain, but this time, the
w
−w
Nyquist rate is not met, and the copies of the frequency response
0
w
s
s
overlap. In such a case, the signal is said to be aliased.
Fs(w)
Not surprisingly, the Nyquist rate depends on both the character-
w
istics of the continuous-time signal, and the sampling rate. More
Figure 3.4 Sampling:
precisely, as the maximum frequency, ωn, of the continuous-time
aliasing
signal increases, the sampling period, T must be made smaller
(i.e., denser sampling), which in turn increases ωs, preventing
overlap of the frequency responses. In other words, a signal that
changes slowly and smoothly can be sampled fairly coarsely, while
a signal that changes quickly requires more dense sampling.
3.3 Discrete to Continuous
If the Nyquist rate is met, then a discrete-time signal fully charac-
terizes the continuous-time signal from which it was sampled. On
the other hand, if the Nyquist rate is not met, then the sampling
leads to aliasing, and the discrete-time signal does not accurately
represent its continuous-time counterpart. In the former case, it
30
is possible to reconstruct the original continuous-time signal, from
the discrete-time signal. In particular since the frequency response
of the discrete-time signal contains exact copies of the original
continuous-time signals frequency response, we need only extract
one of these copies, and inverse transform the result. The result
will be identical to the original signal. In order to extract a single
copy, the Fourier transform of the sampled signal is multiplied by
Fs(w)
an ideal reconstruction filter as shown in Figure 3.5. This filter has
unit value between the frequencies −π/T to π/T and is zero else-
w
where. This frequency band is guaranteed to be greater than the
pi/T
Nyquist frequency, ωn (i.e., ωs = 2π/T > 2ωn, so that π/T > ωn).
In the space domain, this ideal reconstruction filter has the form:
Figure 3.5
Reconstruction
sin(πx/T )
h(x) =
,
(3.6)
πx/T
1
and is often referred to as the ideal sync function. Since recon-
struction in the frequency domain is accomplished by multipli-
cation with the ideal reconstruction filter, we could equivalently
0
reconstruct the signal by convolving with the ideal sync in the
space domain.
0
Figure 3.6 Ideal sync
Example 3.1
Consider the following continuous-time signal:
f (x)
=
cos(ω0x),
a sinusoid with frequency ω0. We will eventually be interested
in sampling this function and seeing how the effects of aliasing
are manifested. But first, let’s compute the Fourier transform of
this signal:
∞
F (ω)
=
f (k)e−iωk
k=−∞
∞
=
cos(ω0k)(cos(ωk) − i sin(ωk))
k=−∞
∞
=
cos(ω0k) cos(ωk) − i cos(ω0k) sin(ωk)
k=−∞
First let’s consider the product of two cosines. It is easy to show
from basic trigonometric identities that cos(A) cos(B) = 0 when
A = B, and is equal to π when |A| = |B|. Similarly, one can
show that cos(A) sin(B) = 0 for all A and B. So, the Fourier
transform of cos(ω0x) = π for |ω| = ω0, and is 0 otherwise (see
below). If the sampling rate is greater than 2ω0, then there will
be no aliasing, but if the sampling rate is less than 2ω0, then
the reconstructed signal will be of the form cos((ωs − ω0)x), that
is, the reconstructed signal will be appear as a lower frequency
sinusoid - it will be aliased.
31
F(w)
w
−w 0
w 0
Sampling
No Aliasing
Aliasing
Fs(w)
Fs(w)
w
w
−w 0
w 0
−w 0
w 0
We will close this chapter by drawing on the linear algebraic frame-
work for additional intuition on the sampling and reconstruction
process. First we will need to restrict ourselves to the sampling of
an already sampled signal, that is, consider a m-dimensional sig-
nal sub-sampled to a n-dimensional signal. We may express this
operation in matrix form as follows:
1
0 0 0
. . .
0 0 0
0
f1
g
1
0
0 1 0
. . .
0 0 0
0
f2
.
.
.
.
.
.. = ..
. .
.
. ..
g
n
0 0 0 0 ... 0 1 0 0
fm−1
1
0 0 0
. . .
0 0 0
1
fm
gn = Sn×mfm,
(3.7)
where the subscripts denote the vector and matrix dimensions,
and in this example n = m/2. Our goal now is to determine when
it is possible to reconstruct the signal f , from the sub-sampled
signal g. The Nyquist sampling theory tells us that if a signal is
band-limited (i.e., can be written as a sum of a finite number of
sinusoids), then we can sample it without loss of information. We
can express this constraint in matrix notation:
fm = Bm×nwn,
(3.8)
where the columns of the matrix B contains the basis set of si-
nusoids - in this case the first n sinusoids. Substituting into the
above sampling equation gives:
gn = Sn×mBm×nwn
= Mn×nwn.
(3.9)
If the matrix M is invertible, then the original weights (i.e., the
representation of the original signal) can be determined by sim-
ply left-multiplying the sub-sampled signal g by M −1. In other
32
words, Nyquist sampling theory can be thought of as simply a
matrix inversion problem. This should not be at all surprising,
the trick to sampling and perfect reconstruction is to simply limit
the dimensionality of the signal to at most twice the number of
samples.
33
4. Digital Filter Design
Recall that the class of linear time-invariant systems are fully char-
4.1 Choosing
a
acterized by their impulse response. More specifically, the output
Frequency Re-
of a linear time-invariant system to any input f [x] can be deter-
sponse
mined via a convolution with the impulse response h[x]:
4.2 Frequency
g[x] = f [x] ⋆ h[x].
(4.1)
Sampling
4.3 Least-Squares
Therefore the filter h[x] and the linear-time invariant system are
synonymous. In the frequency domain, this expression takes on
4.4 Weighted
the form:
Least-Squares
G[ω] = F [ω]H[ω].
(4.2)
In other words, a filter modifies the frequencies of the input signal.
It is often the case that such filters pass certain frequencies and
attenuate others (e.g., a lowpass, bandpass, or highpass filters).
The design of such filters consists of four basic steps:
1. choose the desired frequency response
2. choose the length of the filter
3. define an error function to be minimized
4. choose a minimization technique and solve
The choice of frequency response depends, of course, on the design-
ers particular application, and its selection is left to their discre-
tion. We will however provide some general guidelines for choosing
a frequency response that is amenable to a successful design. In
choosing a filter size there are two conflicting goals, a large filter al-
lows for a more accurate match to the desired frequency response,
however a small filter is desirable in order to minimize computa-
tional demands 4. The designer should experiment with varying
size filters until an equitable balance is found. With a frequency
response and filter size in hand, this chapter will provide the com-
putational framework for realizing a finite length filter that “best”
approximates the specified frequency response. Although there are
numerous techniques for the design of digital filters we will cover
only two such techniques chosen for their simplicity and generally
good performance (see Digital Filter Design by T.W. Parks and
C.S. Burrus for a full coverage of many other approaches).
4In multi-dimensional filter design, separability is also a desirable property.
34
4.1 Choosing a Frequency Response
A common class of filters are bandpass in nature, that is, they pass
| H(w) |
certain frequencies and attenuate others. An ideal lowpass, band-
pass, and highpass filter are illustrated in Figure 4.1 Shown is the
Passband
magnitude of the frequency response in the range [0, π], since we
Stopband
are typically interested in designing real-valued, linear-phase fil-
w
0
pi
ters, we need only specify one-half of the magnitude spectrum (the
response is symmetric about the origin). The responses shown in
Figure 4.1 are often referred to as brick wall filters because of their
abrupt fall-off. A finite-length realization of such a filter produces
undesirable “ringing” known as Gibbs phenomena as shown below
in the magnitude of the frequency response of ideal lowpass filters
of length 64, 32, and 16 (commonly referred to as the filter tap
size).
Figure 4.1
64 taps
32 taps
16 taps
Ideal lowpass, bandpass,
1
1
1
and highpass
0.5
0.5
0.5
0
0
0
0
pi/2
pi
0
pi/2
pi
0
pi/2
pi
| H(w) |
These effects are particularly undesirable because neighboring fre-
quencies may be passed or attenuated by wildly varying amounts,
leading to general instabilities. To counter this problem, the de-
w
0
pi
signer is resigned to sacrificing the ideal response for a “softer”
frequency response, Figure 4.2.
Such a frequency response is
amenable to a small finite-length filter free of significant ringing
artifacts. The specific functional form of the soft falloff is some-
what arbitrary, however one popular form is a raised cosine. In
its most general form, the frequency response takes the follow-
ing form, where the bandpass nature of the response is controlled
through ω0, ω1, ∆ω0, and ∆ω1.
Figure 4.2 Soft lowpass,
0,
ω < ω0
1
bandpass, and highpass
[1
2
− cos(π(ω − ω0)/∆ω0)],
ω0 ≤ ω < ω0 + ∆ω0
H(ω) =
1,
ω0 + ∆ω0
1
≤ ω < ω1 − ∆ω1
[1+cos(π(ω
2
− (ω1 − ∆ω1))/∆ω1)] , ω1 − ∆ω1 ≤ ω < ω1
0,
ω1
∆ω
∆ω
≤ ω.
1
0
1
In general,
a larger transition width (i.e., ∆ω0, and ∆ω1) allows
0.5
for smaller filters with minimal ringing artifacts. The tradeoff,
ω
ω
0
1
of course, is that a larger transition width moves the frequency
0
response further from the ideal brick-wall response. In specifying
0
pi
only half the frequency response (from [0, π]) we are implicitly
Figure 4.3 Frequency
imposing a symmetric frequency response and thus assuming that
response
the desired filter is symmetric.
35
4.2 Frequency Sampling
Once the desired frequency response has been chosen, the more
difficult task of designing a finite-length filter that closely approx-
imates this response begins. In general this problem is hard be-
cause by restricting ourselves to a small finite-length filter we are
in effect asking to fit an arbitrarily complex function (the desired
frequency response) with the sum of a small number of sinusoids.
We begin with the most straight-forward design method - fre-
quency sampling. Our goal is to design a filter h whose Fourier
transform best approximates the specified response H, that is:
F(h) = H
(4.3)
where F is the Fourier transform operator. By applying the inverse
Fourier transform we obtain a solution:
F−1(F(h)) = F−1(H)
h = F−1(H).
(4.4)
In other words, the design simply involves inverse Fourier trans-
forming the specified frequency response. This series of steps can
be made more explicit and practical for computer implementation
by expressing the initial constraint (Equation (4.3)) in matrix no-
tation:
M h = H
(4.5)
where H is the n-dimensional sampled frequency response, M is
the n × n Fourier matrix (Equation (2.28)), and n is the chosen
filter size. The filter h can be solved for by left multiplying both
sides of Equation (4.5) by the inverse of the matrix F :
h = M −1H.
(4.6)
Since the matrix M is square, this design is equivalent to solv-
ing for n unknowns (the filter taps) from n linearly independent
equations. This fact illustrates the shortcomings of this approach,
namely, that this method produces a filter with a frequency re-
sponse that exactly matches the sampled response, but places no
constraints on the response between sampled points. This restric-
tion can often lead to poor results that are partially alleviated by
the least-squares design presented next.
36
4.3 Least-Squares
Our goal is to design a filter h that “best” approximates a specified
frequency response. As before this constraint can be expressed as:
M h = H,
(4.7)
where M is the N × n Fourier matrix (Equation (2.28)), H is the
N sampled frequency response, and the filter size is n. Note that
unlike before, this equation is over constrained, having n unknowns
in N > n equations. We can solve this system of equations in a
least-squares sense by first writing a squared error function to be
minimized:
E(h) = | Mh − H |2
(4.8)
In order to minimize 5, we differentiate with respect to the un-
known h:
dE(h)
= 2M t| Mh − H |
dh
= 2M tM h − 2MtH,
(4.9)
then set equal to zero, and solve:
h = (M tM )−1M tH
(4.10)
1
Shown in Figure 4.4 are a set of lowpass, bandpass, and highpass
0.5
16-tap filters designed using this technique. In this design, the
frequency response was of the form given in Equation (4.3), with
0
start and stop bands of [ω0, ω1] = [0, π/2], [π/4, 3π/4], and [π/2, π],
0
pi/2
pi
and a transition width of ∆ω0 = ∆ω1 = π/4. The frequency
1
response was sampled at a rate of N = 512.
0.5
Finally, note that the previous frequency sampling design is equiv-
alent to the least-squares design when the sampling of the Fourier
00
pi/2
pi
basis is the same as the filter size (i.e., N = n).
1
4.4 Weighted Least-Squares
0.5
One drawback of the least-squares method is that the error func-
tion (Equation (4.8)) is uniform across all frequencies. This is
0
easily rectified by introducing a weighting on the least-squares er-
0
pi/2
pi
ror function:
Figure 4.4 Least-
squares: lowpass, band-
E(h) = W | Mh − H |2
(4.11)
pass, and highpass
5Because of Parseval’s theorem (which amounts to the orthonormality of
the Fourier transform), the minimal error in the frequency domain equates to
a minimal error in the space domain.
37
where W is a diagonal weighting matrix. That is, the diagonal
of the matrix contains the desired weighting of the error across
frequency. As before, we minimize by differentiating with respect
to h:
dE(h)
= 2M tW | Mh − H |
dh
= 2M tW M h − 2WMtH,
(4.12)
then set equal to zero, and solve:
h = (M tW M )−1M tW H.
(4.13)
1
Note that this solution will be equivalent to the original least-
0.5
squares solution (Equation (4.10)) when W is the identity matrix
(i.e., uniform weighting).
00
pi/2
pi
Shown in Figure 4.5 is a comparison of an 8-tap lowpass filter
1
designed with a uniform weighting, and with a weighting that
emphasizes the errors in the low frequency range, W (ω) =
1
.
0.5
(|ω|+1)8
Note that in the case of the later, the errors in the low frequencies
are smaller, while the errors in the high frequencies have increased.
00
pi/2
pi
Figure 4.5
Least-squares
and
weighted least squares
38
5. Photons to Pixels
5.1 Pinhole Cam-
5.1 Pinhole Camera
era
5.2 Lenses
The history of the pinhole camera (or camera obscura) dates back
as early as the fifth century B.C., and continues to be popular to-
5.3 CCD
day among students, artists, and scientists. The Chinese philoso-
pher Mo Ti is believed to be the first to notice that objects reflect
light in all directions and that the light rays that pass through
a small hole produce an inverted image. In its simplest form a
pinhole camera is a light-tight box with a tiny hole in one end and
a photo-sensitive material on the other. Remarkably, this simple
device is capable of producing a photograph. However, the pinhole
camera is not a particularly efficient imaging system (often requir-
ing exposure times as long as several hours) and is more popular
for its artistic value than for its practical value. Nevertheless, the
pinhole camera is convenient because it affords a simple model of
Figure 5.1 Pinhole im-
more complex imaging systems. That is, with a pinhole camera
age formation
model, the projection of points from the three-dimensional world
onto the two-dimensional sensor takes on a particularly simple
form.
Denote a point in the three-dimensional world as a column vector,
P = ( X
Y
Z )t and the projection of this point onto the two
dimensional image plane as p = ( x
y )t. Note that the world
and image points are expressed with respect to their own coordi-
nate systems, and for convenience, the image coordinate system
is chosen to be orthogonal to the Z-axis, i.e., the origins of the
two systems are related by a one-dimensional translation along
Y
y
the Z−axis or optical axis. It is straight-forward to show from a
P
similar triangles argument that the relationship between the world
Z
and image point is:
p
x
X
dX
dY
x = −
and
y =
,
(5.1)
Z
− Z
Figure 5.2 Perspective
projection
where d is the displacement of the image plane along the Z-axis 6
These equations are frequently referred to as the perspective pro-
jection equations. Although non-linear in their nature, the per-
spective projection equations may be expressed in matrix form
6The value d in Equation (5.1) is often referred to as the focal length. We
do not adopt this convention primarily because it is a misnomer, under the
pinhole model all points are imaged in perfect focus.
39
using the homogeneous equations:
X
xs
−d 0 0 0 Y
ys
0
−d 0 0
,
(5.2)
Z
s = 0 0 1 0
1
where the final image coordinates are given by ( x
y )t = ( xs
ys )t.
s
s
An approximation to the above perspective projection equations
is orthographic projection, where light rays are assumed to travel
from a point in the world parallel to the optical axis until they
intersect the image plane. Unlike the pinhole camera and perspec-
tive projection equations, this model is not physically realizable
and is used primarily because the projection equations take on a
particularly simple linear form:
Y
y
x = X
and
y = Y.
(5.3)
P
p
And in matrix form:
Z
x
X
X
x
1 0
0
=
Y
y
0 1
0
Figure 5.3 Orthographic
Z
(5.4)
projection
Orthographic projection is a reasonable approximation to perspec-
tive projection when the difference in depth between points in the
world is small relative to their distance to the image plane. In
the special case when all the points lie on a single frontal-parallel
surface relative to the image plane (i.e., d is constant in Equa-
Z
tion (5.1)), the difference between perspective and orthographic is
only a scale factor.
5.2 Lenses
It is important to remember that both the perspective and or-
thographic projection equations are only approximations of more
complex imaging systems. Commercial cameras are constructed
with a variety of lenses that collect and focus light onto the im-
age plane. That is, light emanates from a point in the world
in all directions and, whereas a pinhole camera captures a single
Y
light ray, a lens collects a multitude of light rays and focuses the
y
P
light to a small region on the image plane. Such complex imaging
Z
systems are often described with the simpler thin-lens model. Un-
der the thin-lens model the projection of the central or principal
ray obeys the rules of perspective projection, Equation (5.1): the
point P = ( X
Y
Z )t is projected onto the image plane cen-
Figure 5.4 Thin lens
tered about the point ( x
y )t = ( −dX
−dY )t. If the point P
Z
Z
40
is in perfect focus, then the remaining light rays captured by the
lens also strike the image plane at the point p. A point is imaged
in perfect focus if its distance from the lens satisfies the following
thin-lens equation:
1
1
1
+
=
,
(5.5)
Z
d
f
where d is the distance between the lens and image plane along the
optical axis, and f is the focal length of the lens. The focal length
is defined to be the distance from the lens to the image plane such
that the image of an object that is infinitely far away is imaged
in perfect focus. Points at a depth of Zo = Z are imaged onto a
small region on the image plane, often modeled as a blurred circle
with radius r:
R
1
1
1
r =
,
(5.6)
1
f − Z
− d
f − 1
Z
o
o
where R is the radius of the lens. Note that when the depth of a
point satisfies Equation (5.5), the blur radius is zero. Note also
that as the lens radius R approaches 0 (i.e., a pinhole camera),
the blur radius also approaches zero for all points independent of
its depth (referred to as an infinite depth of field).
Alternatively, the projection of each light ray can be described in
the following more compact matrix notation:
l2
1
0
l
=
1
,
(5.7)
α
n2−n1
n1
2
− 1R n
α
2
n2
1
where R is the radius of the lens, n1 and n2 are the index of
refraction for air and the lens material, respectively. l1 and l2 are
the height at which a light ray enters and exits the lens (the thin
lens idealization ensures that l1 = l2). α1 is the angle between the
entering light ray and the optical axis, and α2 is the angle between
the exiting light ray and the optical axis. This formulation is
particularly convenient because a variety of lenses can be described
in matrix form so that a complex lens train can then be modeled
as a simple product of matrices.
Y
y
Image formation, independent of the particular model, is a three-
P1 P2 P
dimensional to two-dimensional transformation. Inherent to such
3
Z
a transformation is a loss of information, in this case depth infor-
p
mation. Specifically, all points of the form Pc = ( cX cY
cZ )t,
x
X
for any c ∈ R, are projected to the same point (x y )t - the pro-
jection is not one-to-one and thus not invertible. In addition to
Figure 5.5
this geometric argument for the non-invertibility of image forma-
Non-invertible projection
tion, a similarly straight-forward linear algebraic argument holds.
41
In particular, we have seen that the image formation equations
may be written in matrix form as, p = Mn×mP , where n < m
(e.g., Equation (5.2)). Since the projection is from a higher di-
mensional space to a lower dimensional space, the matrix M is
not invertible and thus the projection is not invertible.
5.3 CCD
To this point we have described the geometry of image formation,
how light travels through an imaging system. To complete the im-
age formation process we need to discuss how the light that strikes
the image plane is recorded and converted into a digital image.
The core technology used by most digital cameras is the charge-
coupled device (CCD), first introduced in 1969. A basic CCD
consists of a series of closely spaced metal-oxide-semiconductor
capacitors (MOS), each one corresponding to a single image pixel.
In its most basic form a CCD is a charge storage and transport de-
vice: charge is stored on the MOS capacitors and then transported
across these capacitors for readout and subsequent transformation
to a digital image. More specifically, when a positive voltage, V , is
applied to the surface of a P-type MOS capacitor, positive charge
migrates toward ground. The region depleted of positive charge
is called the depletion region. When photons (i.e., light) enter the
Light
V
depletion region, the electrons released are stored in this region.
The value of the stored charge is proportional to the intensity of
the light striking the capacitor. A digital image is subsequently
Depletion
region
formed by transferring the stored charge from one depletion re-
gion to the next. The stored charge is transferred across a series
of MOS capacitors (e.g., a row or column of the CCD array) by
MOS
sequentially applying voltage to each MOS capacitor. As charge
Ground
passes through the last capacitor in the series, an amplifier con-
verts the charge into a voltage. An analog-to-digital converter
Figure 5.6 MOS capaci-
then translates this voltage into a number (i.e., the intensity of an
tor
image pixel).
42
6. Point-Wise Operations
6.1 Lookup Table
6.1 Lookup Table
The internal representation of a digital image is simply a matrix
6.2 Brightness
of numbers representing grayscale or color values. But when an
/Contrast
image is displayed on a computer monitor we typically do not see
6.3 Gamma
a direct mapping of the image. An image is first passed through a
Correction
lookup table (LUT) that maps the image intensity values to bright-
ness values, Figure 6.1.
If the lookup table is linear with unit
6.4 Quantize
slope and zero intercept then the image is directly mapped to the
/Threshold
display, otherwise, the displayed image will not be an exact rep-
6.5 Histogram
resentation of the underlying image. For example, most computer
Equalize
monitors intentionally impose a non-linear LUT of the general
form D = Iα (i.e., gamma correction), where α is a real num-
ber, and D and I are the displayed and image values. A variety
white
of interesting visual effects can be achieved by simple manipula-
tions of the functional form of the LUT. Keep in mind though
that in manipulating the lookup table, the underlying image is
left untouched, it is only the mapping from pixel value to display
brightness that is being effected.
black
0
255
6.2 Brightness/Contrast
Figure 6.1 Lookup table
Perhaps the most common and familiar example of a LUT ma-
nipulation is to control the brightness or darkness of an image as
white
shown in Figure 6.3. The bright and dark images of Einstein were
created by passing the middle image through the LUTs shown in
the same figure. The functional form of the LUT is a unit-slope
line with varying intercepts: g(u) = u + b, with the image inten-
sity values u ∈ [0,1]. A value of b > 0 results in a brightening
black
of the image and b < 0 a darkening of the image. Another com-
min
max
mon manipulation is that of controlling the contrast of an image
as shown in Figure 6.4. The top image is said to be high contrast
Figure 6.2 Autoscale
and the bottom image low contrast, with the corresponding LUTs
shown in the same figure. The functional form of these LUTs is
linear: g(u) = mu + b, where the relationship between the slope
and intercept is b = 1/2(1 − m). The image contrast is increased
with a large slope and negative intercept (in the limit, m → ∞
and b → −∞), and the contrast is reduced with a small slope and
positive intercept (m → 0 and b → 1/2). The image is inverted
with m = −1 and b = 1, that is white is mapped to black, and
black is mapped to white. Of course, an image can be both con-
trast enhanced and brightened or darkened by simply passing the
image through two (or more) LUTs.
43
Autoscaling is a special case of contrast enhancement where the
minimum image intensity value is mapped to black and the maxi-
mum value is mapped to white, Figure 6.2. Autoscaling maximizes
the contrast without saturating at black or white. The problem
with this sort of autoscaling is that a few stray pixels can dictate
the contrast resulting in a low-contrast image. A less sensitive ap-
proach is to sort the intensity values in increasing order and map
the 1% and 99% intensity values to black and white, respectively.
Although this will lead to a small amount of saturation, it rarely
fails to produce a high-contrast image.
6.3 Gamma Correction
Typically, high contrast images are visually more appealing. How-
ever a drawback of linear contrast enhancement described above is
that it leads to saturation at both the low and high end of the in-
tensity range. This may be avoided by employing a non-linear con-
Bright
trast adjustment scheme, also realizable as a lookup table (LUT)
manipulation. The most standard approach is gamma correction,
where the LUT takes the form:
g(u) = uα,
(6.1)
Dark
Figure 6.3 Brightness
where α > 1 increases contrast, and α < 1 reduces contrast.
Shown in Figure 6.5 are contrast enhanced (top: α = 2) and
contrast reduced (bottom: α = 1/2) images. Note that with the
intensity values u scaled into the range [0, 1], black (0) and white
(1) are mapped to themselves. That is, there is no saturation
at the low or high end. Gamma correction is widely used in a
number of devices because it yields reasonable results and is eas-
ily parameterized. One drawback to this scheme is that the gray
values are mapped in an asymmetric fashion with respect to mid-
level gray (0.5). This may be alleviated by employing a sigmoidal
non-linearity of the form
1
g(u) =
.
(6.2)
1 + e−αu+β
In order that g(u) be bound by the interval [0, 1], it must be
scaled as follows: (g(u) − c1)/c2, where c1 = 1/(1 + eβ) and
c2 = 1/(1+e−α+β)−c1. This non-linearity, with its two degrees of
freedom, is more versatile and can produce a more balanced con-
High
trast enhancement. Shown in Figure 6.6 is a contrast enhanced
image with α = 12 and β = 6.
Low
Figure 6.4 Contrast
44
6.4 Quantize/Threshold
A digital image, by its very nature, is quantized to a discrete num-
ber of intensity values. For example an image quantized to 8-bits
contains 28 = 256 possible intensity values, typically in the range
[0, 255]. An image can be further quantized to a lower number of
bits (b) or intensity values (2b). Quantization can be accomplished
by passing the image through a LUT containing a step function,
Figure 6.7, where the number of steps governs the number of inten-
sity values. Shown in Figure 6.7 is an image of Einstein quantized
to five intensity values, notice that all the subtle variations in the
curtain and in his face and jacket have been eliminated. In the
limit, when an image is quantized to one bit or two intensity val-
ues, the image is said to be thresholded. Shown in Figure 6.8 is a
thresholded image of Einstein and the corresponding LUT, a two-
step function. The point at which the step function transitions
from zero to one is called the threshold and can of course be made
to be any value (i.e., slid left or right).
Figure 6.5 Contrast:
6.5 Histogram Equalize
Gamma
The intensity values of a typical image are often distributed un-
evenly across the full range of 0 to 255 (for an 8-bit image), with
most the mass near mid-gray (128) and falling off on either side,
Figure 6.9. An image can be transformed so that the distribution
of intensity values is flat, that is, each intensity value is equally
represented in the image. This process is known has histogram
equalization 7. Although it may not be immediately obvious an
image is histogram equalized by passing it through a LUT with
the functional form of the cumulative distribution function. More
specifically, define N (u) as the number of pixels with intensity
value u, this is the image histogram and a discrete approximation
to the probability distribution function. Then, the cumulative
distribution function is defined as:
Figure 6.6 Contrast:
u
Sigmoid
C(y) =
N (i),
(6.3)
i=0
that is, C(u) is the number of pixels with intensity value less than
or equal to u. Histogram equalization then amounts to simply in-
serting this function into the LUT. Shown in Figure 6.9 is Einstein
before and after histogram equalization. Notice that the effect is
similar to contrast enhancement, which intuitively should make
sense since we increased the number of black and white pixels.
7Why anyone would want to histogram equalize an image is a mystery to
me, but here it is in case you do.
45
In all of these examples the appearance of an image was altered by
simply manipulating the LUT, the mapping from image intensity
value to display brightness value. Such a manipulation leaves the
image content intact, it is a non-destructive operation and thus
completely invertible. These operations can be made destructive
by applying the LUT operation directly to the image. For example
an image can be brightened by adding a constant to each pixel,
and then displaying with a linear LUT. Since such an operation
is destructive it may not be inveritble, for example when bright-
ening an 8-bit image, all pixels that exceed the value 255 will be
truncated to 255.
Figure 6.7 Quantize
Figure 6.8 Threshold
15000
10000
5000
00
100
200
15000
10000
5000
00
100
200
46
Figure 6.9 Histogram
equalize
7. Linear Filtering
7.1 Convolution
7.1 Convolution
The one-dimensional convolution sum, Equation (2.5), formed the
7.2 Derivative Fil-
basis for much of our discussion on discrete-time signals and sys-
ters
tems. Similarly the two-dimensional convolution sum will form
7.3 Steerable
the basis from which we begin our discussion on image processing
Filters
and computer vision.
7.4 Edge Detection
The 1-D convolution sum extends naturally to higher dimensions.
Consider an image f [x, y] and a two-dimensional filter h[x, y]. The
7.5 Wiener Filter
2-D convolution sum is then given by:
∞
∞
ωy
g[x, y] =
f [k, l]h[x − k,y − l].
(7.1)
k=−∞ l=−∞
In 1-D, the intuition for a convolution is that of computing inner
ωx
products between the filter and signal as the filter “slides” across
the signal. The same intuition holds in 2-D. Inner products are
computed between the 2-D filter and underlying image as the filter
slides from left-to-right/top-to-bottom.
Figure 7.1 2-D
Frequency
In the Fourier domain, this convolution is equivalent to multiplying
the, now 2-D, Fourier transforms of the filter and image, where the
2-D Fourier transform is given by:
∞
∞
F [ωx, ωy] =
f [k, l]e−i(ωxk+ωyl).
(7.2)
k=−∞ l=−∞
The notion of low-pass, band-pass, and high-pass filtering extends
naturally to two-dimensional images. Shown in Figure 7.1 is a
simplified decomposition of the 2-D Fourier domain parameterized
by ωx and ωy ∈ [−π,π]. The inner disc corresponds to the lowest
frequencies, the center annulus to the middle (band) frequencies,
and the outer dark area to the highest frequencies.
Two of the most common (and opposing) linear filtering opera-
tions are blurring and sharpening. Both of these operations can
be accomplished with a 2-D filter and 2-D convolution, or more
efficiently with a 1-D filter and a pair of 1-D horizontal and ver-
tical convolutions. For example, a 2-D convolution with the blur
filter:
0.0625 0.1250 0.0625
0.1250 0.2500 0.1250
0.0625
0.1250
0.0625
47
Figure 7.2 Low-, Band-,
High-pass
can be realized by convolving in the horizontal and vertical direc-
tions with the 1-D filter:
blur = ( 0.25
0.50
0.25 ) .
(7.3)
That is, an outer-product of the 1-D filter with itself yields the
2-D filter - the filters are xy-separable. The separability of 2-D
filters is attractive for two reasons: (1) it is computationally more
efficient and (2) it simplifies the filter design. A generic blur filter
may be constructed from any row of the binomial coefficients:
1
1
1
2
1
1
3
3
1
1
4
6
4
1
1
5
10
10
5
1
1
6
15
20
15
6
1
where each row (filter) should be normalized by it’s sum (i.e., blur
filters should always be unit-sum so as not to increase or decrease
the mean image intensity). The amount of blur is then directly
proportional to the size of the filter. Blurring simply reduces
the high-frequency content in an image. The opposing operation,
sharpening, is meant to enhance the high-frequencies. A generic
separable sharpening filter is of the form:
sharp = ( 0.08
−1.00 0.08 ).
(7.4)
This filter leaves the low-frequencies intact while enhancing the
contribution of the high-frequencies. Shown in Figure 7.3 are re-
sults from blurring and sharpening.
7.2 Derivative Filters
Discrete differentiation forms the foundation for many applications
Figure 7.3 Blur
and
in image processing and computer vision. We are all familiar with
Sharpen
the definition of the derivative of a continuous signal f (x):
f (x + ε)
D{f(x)} = lim
− f(x).
(7.5)
ε→0
ε
This definition requires that the signal f (x) be well defined for all
x ∈ R. So, does it make sense to differentiate a discretely sampled
signal, D{f[x]}, which is only defined over an integer sampling
lattice? Strictly speaking, no. But our intuition may be that this
is not such an unreasonable request. After all, we know how to
differentiate f (x), from which the sampled signal f [x] was derived,
so why not just differentiate the continuous signal f (x) and then
sample the result? Surely this is what we have in mind when we
48
ask for the derivative of a sampled signal. But one should not be
fooled by the seeming simplicity of our intuition, as we will soon
discover the design of an accurate and efficient discrete derivative
operator will prove to be sufficiently challenging.
Recall from earlier chapters that under certain conditions (Nyquist
theory), the relationship between the continuous and sampled sig-
nals can be expressed precisely as:
f (x) = f [x] ⋆ h(x),
(7.6)
where h(x) = sin(πx/T )/(πx/T ) is the ideal sync, and ⋆ is the
convolution operator. Now, applying the continuous differential
operator to both sides yields:
D{f(x)} = D{f[x] ⋆ h(x)},
(7.7)
and expressing the right-hand side in terms of the convolution
sum:
∞
D{f(x)} = D
f [k]h(x − k)
k=−∞
∞
=
f [k]D{h(x − k)}
k=−∞
= f [x] ⋆ D{h(x)}.
(7.8)
Notice that the derivative operator D{·} is being applied only
to continuous entities. Having computed the desired derivatives,
we need only sample the results, denoting S{·} as the sampling
operator:
S
1
{ D{f(x)} } = f[x] ⋆ S{ D{h(x)} }
S{f′(x)} = f[x] ⋆ S{h′(x)}
f ′[x] = f [x] ⋆ h′[x].
(7.9)
0
On the left-hand side of the above equation is the desired quan-
tity, the derivative of the sampled signal. On the right-hand side
0
is a discrete convolution between two known quantities, the sam-
pled derivative of the sync and the original sampled signal. The
derivative of the sync can be expressed analytically by simply dif-
0
ferentiating the sync function:
π2x/T 2 cos(πx/T )
h′(x) =
− π/T sin(πx/T), (7.10)
0
(πx/T )2
Figure 7.4 Ideal
sync
where T is the sampling period at which f (x) was sampled. So,
and its derivative
if the signal f (x) is sampled above the Nyquist rate and if it is in
49
fact differentiable, then Equation (7.9) tells us that we can exactly
compute the derivative of the sampled signal f [x], an altogether
happy ending.
If you are feeling a bit uneasy it is for a good reason. Although
mathematically correct, we have a solution for differentiating a
discretely sampled signal that is physically unrealizable. In partic-
ular the derivative of the sync, h′(x), is spatially infinite in extent,
meaning that it cannot be implemented on a finite machine. And
even worse, h′(x) falls off slowly from the origin so that truncation
will cause significant inaccuracies. So we are going to have to part
with mathematical perfection and design a finite-length filter.
To begin we need to compute the frequency response of the ideal
derivative filter. We can compute the response indirectly by first
expressing f [x] in terms of its Fourier series:
1
π
f [x] =
F [ω]eiωx,
(7.11)
2π ω=−π
and then differentiating both sides with respect to x:
1
π
D{f[x]} =
F [ω]D
2π
{eiωx}
ω=−π
1
π
=
iωF [ω]eiωx.
(7.12)
2π ω=−π
Differentiation in the space domain is then seen to be equivalent to
multiplying the Fourier transform F [ω] by an imaginary ramp iω.
And since multiplication in the frequency domain is equivalent
to convolution in the space domain, an imaginary ramp is the
frequency response of the ideal derivative filter. Trying to directly
design a finite length filter to this response is futile because of the
pi
discontinuity at −π/π, which of course accounts for the spatially
infinite extent of h′(x). So we are resigned to designing a filter with
0
a periodic frequency response that “best” approximates a ramp.
The simplest such approximation is that of a sinusoid where, at
least in the low-frequency range, the match is reasonably good
−pi
−pi
0
pi
(i.e., sin(ω) = ω, for small ω). Employing the least-squares filter
Figure 7.5 Ideal and ap-
design technique (Equation (4.8)) we formulate a quadratic error
proximate derivative fre-
function to be minimized:
quency response
E(h) = | Mh − H |2,
(7.13)
where M is the N × n Fourier matrix (Equation (2.28)), H is the
N sampled desired frequency response, and n the filter size is.
50
To minimize we differentiate, set equal to zero and solve for the
minimal solution:
h = (M tM )−1M tH
(7.14)
Since the desired frequency response, a sinusoid, has only two de-
grees of freedom, amplitude and phase, a 2-tap filter will suffice
(i.e., n = 2). The resulting filter is of the form h = ( 0.5
−0.5).
Intuitively this is exactly what we should have expected - for ex-
ample, applying this filter via a convolution and evaluating at,
arbitrarily, n = 0 yields:
f ′[x] = h[x] ⋆ f [x]
∞
=
h[x − k]f[k]
k=−∞
f ′[0] = h[1]f [−1] + h[0]f[0]
= 0.5f [0] − 0.5f[−1].
(7.15)
Note that the derivative is being approximated with a simple two-
point difference, that is, a discrete approximation to the continu-
ous definition in Equation (7.5). We could of course greatly im-
prove on this filter design. But since we are really interested in
multi-dimensional differentiation, let’s put aside further analysis
of the one-dimensional case and move on to the two-dimensional
case.
It has been the tendency to blindly extend the one-dimensional
design to higher-dimensions, but, as we will see shortly, in higher-
dimensions the story becomes slightly more complicated. In the
context of higher-dimensional signals we first need to consider par-
tial derivatives. For example the partial derivative of a two dimen-
sional signal f (x, y) in it’s first argument is defined as:
∂f (x, y)
fx(x, y) ≡
∂x
f (x + ε, y)
= lim
− f(x,y).
(7.16)
ε→0
ε
According to the Nyquist theory, the continuous and discrete sig-
nals (if properly sampled) are related by the following equality:
f (x, y) = f [x, y] ⋆ h(x, y),
(7.17)
where h(x, y) = sin(πx/T ) sin(πy/T )
π2xy/T 2
is the two-dimensional ideal sync.
As before we apply the continuous partial differential operator to
both sides:
Dx{f(x,y)} = f[x,y] ⋆ Dx{h(x,y)},
(7.18)
51
noting again that the differential operator is only applied to contin-
uous entities. Since the two-dimensional sync is separable (i.e., h(x, y) =
h(x) ⋆ h(y)), the above equation can be rewritten as:
fx(x, y) = f [x, y] ⋆ Dx{h(x) ⋆ h(y)}
= f [x, y] ⋆ Dx{h(x)} ⋆ h(y)
= f [x, y] ⋆ h′(x) ⋆ h(y).
(7.19)
And finally, sampling both sides gives an expression for the partial
derivative of the discretely sampled two-dimensional signal:
S{fx(x,y)} = f[x,y] ⋆ S{h′(x)} ⋆ S{h(y)}
fx[x, y] = f [x, y] ⋆ h′[x] ⋆ h[y].
(7.20)
Notice that calculating the partial derivative requires a pair of
one-dimensional convolutions: a derivative filter, h′[x], in the di-
mension of differentiation, and an interpolation filter, h[y], in
the other dimension (for multi-dimensional signals, all remain-
ing dimensions would be convolved with the interpolation filter).
f’[x]
Since two-dimensional differentiation reduces to a pair of one-
dimensional convolutions it is tempting to simply employ the same
differentiation filter used in the one-dimensional case. But since a
pair of filters are now required perhaps we should give this some
additional thought.
f[y]
In some ways the choice of filters seems trivial: chose an interpo-
lation function h(x), differentiate it to get the derivative function
Figure 7.6 Horizontal
h′(x), and sample these functions to get the final digital filters
partial differentiation
h[x] and h′[x]. So how is this different from the one-dimensional
case? In the one-dimensional case only the derivative filter is em-
ployed, whereas in the two-dimensional case we require the pair
of filters. And by our formulation we know that the pair of fil-
ters should satisfy the relationship that one is the derivative of
the other h′(x) = D(h(x)). And in fact this constraint is au-
tomatically enforced by the very nature in which the continu-
ous functions are chosen, but in the final step, these functions
are sampled to produce discrete filters. This sampling step typi-
cally destroys the required derivative relationship, and, although
a seemingly subtle point, has dramatic effects on the accuracy of
the resulting derivative operator. For example consider the often
used Sobel derivative filters with h[x] = ( 1
√2 1)/(2 + √2)
and h′[x] = ( 1
0 −1)/3. Shown in Figure 7.7 are the magni-
tudes of the Fourier transform of the derivative filter (solid line)
−pi
0
pi
and the interpolation filter times iω (i.e., frequency domain differ-
Figure 7.7
entiation). If the filters obeyed the required derivative relationship
Sobel frequency response
than these curves would be exactly matched, which they clearly
52
are not. The mismatching of the filters results in gross inaccura-
cies in derivative measurements. Let’s see then if we can design a
better set of filters.
We begin by writing down the desired relationship between the
derivative and interpolation filters, most conveniently expressed
in the frequency domain:
H′(ω) = iωH(ω),
(7.21)
from which we can write a weighted least-squares error functional
to be minimized:
E(H, H′) =
dω W (ω)(iωH(ω) − H′(ω)) 2 , (7.22)
where W (ω) is a frequency weighting function. Next, we write a
discrete approximation of this continuous error functional over the
n-vectors h and h′ containing the sampled derivative and interpo-
lation filters, respectively:
E(h, h′) = |W(F′h − Fh′)|2,
(7.23)
where the columns of the matrix Fm×n contain the first n Fourier
basis functions (i.e., a discrete-time Fourier transform), the matrix
F ′m×n = iωFm×n, and Wm×m is a diagonal frequency weighting
matrix. Note that the dimension n is determined by the filter size
and the dimension m is the sampling rate of the continuous Fourier
basis functions, which should be chosen to be sufficiently large to
avoid sampling artifacts. This error function can be expressed
more concisely as:
E(u) = |Mu|2,
(7.24)
where the matrix M and the vector u are constructed by “packing
together” matrices and vectors:
h
M = ( W F ′
| −WF )
and
u =
.
(7.25)
h′
The minimal unit vector u is then simply the minimal-eigenvalue
eigenvector of the matrix M tM . After solving for u, the derivative
and interpolation filters can be “unpacked” and normalized so that
the interpolation filter is unit sum. Below are the resulting filter
values for a 3-tap and 5-tap filter pair.
53
Shown in Figure 7.8 are the matching of these filters in the fre-
quency domain. Notice that the 5-tap filters are nearly perfectly
matched.
h
0.223755 0.552490 0.223755
h′
0.453014
0.0
-0.453014
h
0.036420 0.248972 0.429217 0.248972
0.036420
h′
0.108415 0.280353
0.0
-0.280353 -0.108415
−pi
0
pi
Higher-order derivative filters can be designed by replacing the
initial constraint in Equation (7.21) with H′(ω) = (iω)kH(ω) for
a kth order derivative.
A peculiar aspect of this filter design is that nowhere did we ex-
plicitly try to model a specified frequency response. Rather, the
−pi
0
pi
design fell naturally from the relationship between the continuous-
Figure 7.8
and discrete-time signals and the application of the continuous
Frequency
response
of
derivative operator, and in this way is quite distinct from the one-
matched derivative filters
dimensional case. The proper choice of derivative filters can have
a dramatic impact on the applications which utilize them. For
example, a common application of differential measurements is in
measuring motion from a movie sequence f (x, y, t). The standard
formulation for motion estimation is:
fx(x, y, t)vx(x, y) + fy(x, y, t)vy(x, y) + ft(x, y, t) = 0, (7.26)
where the motion vector is v = ( vx vy )t, and fx(·), fy(·), and
ft(·) are the partial derivatives with respect to space and time.
Shown in Figure 7.9 are the resulting motion fields for a sim-
ple translational motion with the Sobel (top panel) and matched
(bottom) derivative filters used to compute the various derivatives.
Although these filters are the same size, the difference in accuracy
is significant.
Figure 7.9
7.3 Steerable Filters
Differential motion esti-
mation
In the previous section we showed how to compute horizontal and
vertical partial derivatives of images. One may naturally wonder
how to compute a derivative in an arbitrary direction. Quite re-
markably it turns out that we need not design a new set of filters
for each possible direction because the derivative in any direction
can be synthesized from a linear combination of the horizontal and
vertical derivatives. This property of derivatives has been termed
steerability. There are several formulations of this property, we
chose to work in the frequency domain where differentiation takes
on a particularly simple form.
54
To begin, we express a two-dimensional image with respect to its
Fourier series:
π
π
f (x, y) =
F (ωx, ωy)e−j(ωxx+ωyy).
(7.27)
ωx=−π ωy=−π
Differentiating 8 both sides with respect to x gives:
π
π
fx(x, y) =
−jωxF(ωx,ωy)e−j(ωxx+ωyy).(7.28)
ωx=−π ωy=−π
That is, in the frequency domain differentiation in the horizon-
tal direction u is equivalent to multiplying the Fourier transform
F (ωx, ωy) by an imaginary horizontal ramp, −jωx. Similarly, the
vertical partial derivative in v is:
π
π
fy(x, y) =
−jωyF(ωx,ωy)e−j(ωxx+ωyy).(7.29)
ωx=−π ωy=−π
Now, differentiation in the vertical direction v is equivalent to
multiplying the Fourier transform by an imaginary vertical ramp.
This trend generalizes to arbitrary directions, that is, the partial
derivative in any direction α can be computed by multiplying the
Fourier transform by an imaginary oriented ramp −jωα:
π
π
fα(x, y) =
−jωαF(ωx,ωy)e−j(ωxx+ωyy),(7.30)
ωx=−π ωy=−π
where the oriented ramp can be expressed in terms of the horizon-
tal and vertical ramps:
ωα = cos(α)ωx + sin(α)ωy.
(7.31)
Substituting this definition back into the partial derivative in α,
Equation (7.30), gives:
π
π
fα(x, y) =
−j[cos(α)ωx + sin(α)ωy]F(ωx,ωy)e−j(ωxx+ωyy)
ωx=−π ωy=−π
π
π
= cos(α)
−jωxF(ωx,ωy)e−j(ωxx+ωyy)
ωx=−π ωy=−π
π
π
+ sin(α)
−jωyF(ωx,ωy)e−j(ωxx+ωyy)
ωx=−π ωy=−π
= cos(α)fx(x, y) + sin(α)fy(x, y).
(7.32)
8Recall that the derivative of an exponential is an exponential, so that
according to the chain rule, Dx{eax} = aeax.
55
Notice that we obtain the horizontal and vertical derivatives when
α = 0 and α = 90. This equation embodies the principle of steer-
ability - the derivative in any direction α can be synthesized from
a linear combination of the partial horizontal and vertical deriva-
tives, fx(x, y) and fy(x, y). Pause to appreciate how remarkable
this is, a pair of directional derivatives is sufficient to represent an
infinite number of other directional derivatives, i.e., α can take on
any real-valued number.
From the previous section we know how to compute the horizon-
tal and vertical derivatives via convolutions with an interpolation
and derivative filter. To compute any other directional derivative
no more convolutions are required, simply take the appropriate
linear combinations of the horizontal and vertical derivatives as
specified in Equation (7.32). Shown in Figure 7.10 from top to
bottom is a disc f (x, y), its horizontal derivative fx(x, y), its ver-
tical derivative fy(x, y), and its steered derivative f45(x, y), where
the steered derivative was synthesized from the appropriate linear
combinations of the horizontal and vertical derivatives. The ob-
vious benefit of steerability is that the derivative in any direction
can be synthesized with minimal computational costs.
Steerability is not limited to first-order derivatives. Higher-order
derivatives are also steerable; the N th-order derivative is steerable
with a basis set of size N + 1. For example, the second-order
derivative in an arbitrary direction can be synthesized as follows:
fαα = cos2(α)fxx + 2 cos(α) sin(α)fxy + sin2(α)fyy,
(7.33)
where for notational simplicity the spatial arguments (x, y) have
been dropped, and the multiple subscripts denote higher-order
differentiation. Note that three partial derivatives are now needed
to steer the second-order derivative. Similarly, the third-order
derivative can be steered with a basis of size four:
fααα = cos3(α)fxxx + 3 cos2(α) sin(α)fxxy
+3 cos(α) sin2(α)fxyy + sin3(α)fyyy. (7.34)
You may have noticed that the coefficients needed to steer the basis
set look familiar, they are the binomial coefficients that come from
a polynomial expansion. More specifically, as in Equation(7.30)
Figure 7.10 Steerability
the N th-order derivative in the frequency domain is computed by
multiplying the Fourier transform by an imaginary oriented ramp
raised to the N th power, (−jωα)N. Expressing this oriented ramp
in terms of the horizontal and vertical ramps provides the basis
and coefficients needed to steer derivatives of arbitrary order:
(ωα)N = (cos(α)ωx + sin(α)ωy)N .
(7.35)
56
Although presented in the context of derivatives, the principle of
steerability is not limited to derivatives. In the most general case,
a two-dimensional filter f (x, y) is steerable in orientation if it can
be expressed as a polar-separable function, g(r)h(θ), where h(θ) is
band-limited. More specifically, for an arbitrary radial component
g(r), and for h(θ) expressed as:
N
h(θ) =
an cos(nθ) + bn sin(nθ)
(7.36)
n=1
then the filter is steerable with a basis size of 2N .
7.4 Edge Detection
Discrete differentiation forms the foundation for many applica-
tions in computer vision. One such example is edge detection - a
topic that has received an excessive amount of attention, but is
only briefly touched upon here. An edge is loosely defined as an
extended region in the image that undergoes a rapid directional
change in intensity. Differential techniques are the obvious choice
for measuring such changes. A basic edge detector begins by com-
puting first-order spatial derivatives of an image f [x, y]:
fx[x, y] = (f [x, y] ⋆ h′[x]) ⋆ h[y]
(7.37)
fy[x, y] = (f [x, y] ⋆ h[x]) ⋆ h′[y],
(7.38)
where h′[·] and h[·] are the derivative and prefilter defined in Sec-
tion 7.2. The “strength” of an edge at each spatial location is
defined to be the magnitude of the gradient vector ▽[x,y] =
( fx[x, y] fy[x, y] ), defined as:
Figure 7.11 Edges
| ▽ [x,y]| =
f 2x[x, y] + f2y[x, y].
(7.39)
As shown in Figure 7.11, the gradient magnitude is only the begin-
ning of a more involved process (not discussed here) of extracting
and localizing the salient and relevant edges.
7.5 Wiener Filter
For any of a number of reasons a digital signal may become cor-
n
rupted with noise. The introduction of noise into a signal is often
modeled as an additive process, ˆ
s = s+n. The goal of de-noising is
to recover the original signal s from the corrupted signal ˆ
s. Given
a single constraint in two unknowns this problem is equivalent to
s
+
s
my asking you “37 is the sum of two numbers, what are they?”
Lacking clairvoyant powers or knowledge of how the individual
Figure 7.12 Additive
numbers were selected we have little hope of a solution. But by
noise
57
making assumptions regarding the signal and noise characteris-
tics and limiting ourselves to a linear approach, a solution can be
formulated known as the Wiener filter, of famed Mathematician
Norbert Wiener (1894-1964).
Having restricting ourselves to a linear solution, our goal is to
1
design a filter h[x] such that:
s[x] = h[x] ⋆ ˆ
s[x]
= h[x] ⋆ (s[x] + n[x]),
(7.40)
that is, when the filter is convolved with the corrupted signal the
0
−pi
0
pi
original signal is recovered. With this as our goal, we reformulate
this constraint in the frequency domain and construct a quadratic
error functional to be minimized:
E(H(ω)) =
dω [H(ω)(S(ω) + N (ω)) − S(ω)]2. (7.41)
For notational simplicity we drop the frequency parameter ω and
Figure 7.13 Wiener fil-
express the integral with respect to the expected value operator
ter
E{·}:
E(H) = E (H(S + N) − S)2
= E H2(S + N)2 − 2HS(S + N) + S2
= E H2(S2 + 2SN + N2) − 2H(S2 + SN) + S2(7.42)
In order to simplify this expression we can assume that the signal
and noise are statistically independent (i.e., E{SN} = 0), yielding:
E(H) = E H2(S2 + N2) − 2HS2 + S2 .
(7.43)
To minimize, we differentiate:
dE(H)
= 2H(S2 + N 2)
dH
− 2S2,
(7.44)
set equal to zero and solve:
S2(ω)
H(ω) =
.
(7.45)
S2(ω) + N 2(ω)
At an intuitive level this frequency response makes sense - when
the signal is strong and the noise is weak the response is close to
1 (i.e., frequencies are passed), and when the signal is weak and
the noise is strong the response is close to 0 (i.e., frequencies are
Figure 7.14 Einstein
stopped). So we now have an optimal (in the least-squares sense)
plus noise
58
frequency response in terms of the signal and noise characteris-
tics, but of course we don’t typically know what those are. But
we can instantiate them by making assumptions about the general
statistical nature of the signal and noise, for example a common
choice is to assume white noise, N (ω) is constant for all ω, and,
for natural images, to assume that S(ω) = 1/ωp. The frequency
response in the top panel of Figure 7.13 was constructed under
these assumptions. Shown in the bottom panel is a 7-tap filter
derived from a least-squares design. This one-dimensional formu-
lation can easily be extended to two or more dimensions. Shown
in Figure 7.14 from top to bottom, is Einstein, Einstein plus noise,
and the results of applying a 7 × 7 Wiener filter. Note that the
noise levels are reduced but that much of the sharp image struc-
ture has also been lost, which is an unfortunate but expected side
effect given that the Wiener filter is low-pass in nature.
59
8. Non-Linear Filtering
8-1 Median Filter
8.1 Median Filter
8-2 Dithering
Noise may be introduced into an image in a number of different
ways. In the previous section we talked about how to remove
noise that has been introduced in an additive fashion. Here we
look at a different noise model, one where a small number of pix-
els are corrupted due to, for example, a faulty transmission line.
The corrupted pixels randomly take on a value of white or black,
hence the name salt and pepper used to describe such noise pat-
terns (Figure 8.1). Shown in the middle panel of Figure 8.1 is the
disastrous result of applying the solution from the additive noise
model (Wiener filter) to the salt and pepper noise image in the
Salt & Pepper Noise
top panel. Trying to average out the noise in this fashion is equiv-
alent to asking for the average salary of a group of eight graduate
students and Bill Gates. As the income of Gates will skew the av-
erage salary so does each noise pixel when its value is so disparate
from its neighbors. In such cases, the mean is best replaced with
the median, computed by sorting the set of numbers and report-
ing on the value that lies midway. Shown in the bottom panel of
Figure 8.1 is the much improved result of applying a 3 × 3 me-
dian filter to the salt and pepper noise image. More specifically,
Wiener
the center pixel of each 3 × 3 neighborhood is replaced with the
median of the nine pixel values in that neighborhood.
Depending on the density of the noise the median filter may need
to be computed over a larger neighborhood. The tradeoff being
that a larger neighborhood leads to a loss of detail, however this
loss of detail is quite distinct from that of averaging. For example,
shown in Figure 8.2 is the result of applying a 15×15 median filter.
Median
Notice that although many of the internal details have been lost
Figure 8.1 Median filter
the boundary contours (edges) have been retained, this is often
referred to as posterization. This effect could never be achieved
with an averaging filter which would indiscriminately smooth over
all image structures.
Because of the non-linear sorting step of a median filter it cannot
be implemented via a simple convolution and is thus often more
costly to implement. Outside the scope of this presentation there
are a number of tricks for reducing the computational demands of
a median filter.
Figure 8.2 15 × 15 me-
dian filter
60
8.2 Dithering
Dithering is a process by which a digital image with a finite num-
ber of gray levels is made to appear as a continuous-tone image.
For example, shown at the top of Figure 8.3 is an 8-bit (i.e., 256
gray values) grayscale image of Richard P. Feynman. Shown below
are two 1 bit (i.e., 2 gray values) images. The first was produced
by thresholding, and the second by dithering. Although in both
images each pixel takes on only one of two gray values (black or
white), it is clear that the final image quality is critically depen-
dent on the way in which pixels are quantized.
There are numerous dithering algorithms (and even more vari-
ants within each algorithm). Sadly there are few quantitative
metrics for measuring the performance of these algorithms. A
standard and reasonably effective algorithm is a stochastic error
diffusion algorithm based on the Floyd/Steinberg algorithm. The
basic Floyd/Steinberg error diffusion dithering algorithm tries to
exploit local image structure to reduce the effects of quantization.
For simplicity, a 1-bit version of this algorithm is described here,
the algorithm extends naturally to an arbitrary number of gray
levels.
This algorithm operates by scanning through the image, left to
right and top to bottom. At each pixel, the gray value is first
thresholded into “black” or “white”, the difference between the
new pixel value and the original value is then computed and dis-
tributed in a weighted fashion to its neighbors. Typically, the error
is distributed to four neighbors with the following weighting:
1
,
16 ×
• 7
3 5 1
where the • represents the thresholded pixel, and the position of
the weights represent spatial position on a rectangular sampling
Figure 8.3 Thresholding
lattice. Since this algorithm makes only a single pass through
and Dithering
the image, the neighbors receiving a portion of the error must
consist only of those pixels not already visited (i.e., the algorithm
is casual). Note also that since the weights have unit sum, the
error is neither amplified nor reduced. As an example, consider
the quantization of an 8-bit image to a 1-bit image. An 8-bit
image has 255 gray values, so all pixels less than 128 (mid-level
gray) are thresholded to 0 (black), and all values greater than 128
are thresholded to 255 (white). A pixel at position (x, y) with
intensity value 120 is thresholded to 0. The error, 120-0 = 120,
is distributed to four neighbors as follows: (7/16)120 is added to
the pixel at position (x + 1, y), (3/16)120 is added to the pixel at
61
position (x − 1,y + 1), (5/16)120 to pixel (x,y + 1) and (1/16)120
to pixel (x + 1, y + 1). The intuition behind this algorithm is
that when the pixel value of 120 is thresholded to 0 that pixel is
made darker. By propagating this error the surrounding pixels are
brightened making it more likely that they will be thresholded to
something brighter. As a result the local neighborhood maintains
it’s average gray value.
Qualitatively, the error diffusion algorithm reduces the effects of
quantization via simple thresholding. However this algorithm does
introduce correlated artifacts due to the deterministic nature of
the algorithm and scanning order. These problems may be par-
tially alleviated by introducing a stochastic process in a variety
of places. Two possibilities include randomizing the error before
distributing it to its neighbors (e.g., randomly scaling the error
in the range 0.9 to 1.1), and alternating the scanning direction
(e.g., odd lines are scanned left to right, and even lines from right
to left).
62
9. Multi-Scale Transforms
63
10. Motion Estimation
10-1 Differential
10.1 Differential Motion
Motion
10-2 Differential
Our visual world is inherently dynamic. People, cars, dogs, etc.
Stereo
are (usually) moving. These may be gross motions, walking across
the room, or smaller motions, scratching behind your ear. Our
task is to estimate such image motions from two or more images
taken at different instances in time.
With respect to notation, an image is denoted as f (x, y) and an
image sequence is denoted as f (x(t), y(t), t), where x(t) and y(t)
are the spatial parameters and t is the temporal parameter. For
example, a sequence of N images taken in rapid succession may be
represented as f (x(t), y(t), t + i∆t) with i ∈ [0,N −1], and ∆t rep-
resenting the amount of time between image capture (typically on
the order of 1/30th of a second). Given such an image sequence,
our task is to estimate the amount of motion at each point in the
image. For a given instant in space and time, we require an esti-
mate of motion (velocity) v = ( vx vy ), where vx and vy denote
the horizontal and vertical components of the velocity vector v.
Shown in Figure 10.1 are a pair of images taken at two moments
in time as a textured square is translating uniformly across the
image. Also shown is the corresponding estimate of motion often
referred to as a flow field. The flow field consists of a velocity vec-
tor at each point in the image (shown of course are only a subset
of these vectors).
In order to estimate motion, an assumption of brightness constancy
is made. That is, it is assumed that as a small surface patch is
moving, its brightness value remains unchanged. This constraint
can be expressed with the following partial differential equation:
Figure 10.1 Flow field
∂f (x(t), y(t), t)
= 0.
(10.1)
∂t
This constraint holds for each point in space and time. Expanding
this constraint according to the chain rule yields:
∂f ∂x
∂f ∂y
∂f
+
+
= 0,
(10.2)
∂x ∂t
∂y ∂t
∂t
where the partials of the spatial parameters x and y with respect
to time correspond to the velocity components:
fxvx + fyvy + ft = 0.
(10.3)
64
The subscripts on the function f denote partial derivatives. Note
again that this constraint holds for each point in space and time
but that for notational simplicity the spatial/temporal parameters
are dropped. This transformed brightness constancy constraint is
rewritten by packing together the partial derivatives and velocity
components into row and column vectors.
v
( f
x
x
fy )
+ f
v
t
= 0
y
f tsv + ft = 0.
(10.4)
The space/time derivatives fs and ft are measured quantities, leav-
ing us with a single constraint in two unknowns (the two compo-
nents of the velocity vector, v). The constraint can be solved by
assuming that the motion is locally similar, and integrating this
constraint over a local image neighborhood. A least-squares error
function takes the form:
2
E(v) =
f tsv +
ft
,
(10.5)
x,y
x,y
To solve for the motion this error function is first differentiated
∂E(v)
= 2
f
∂v
s
f tsv +
ft
= 2
fsf tsv + 2
fsft.
(10.6)
Setting equal to zero and recombining the terms into matrix form
yields:
fx
v
f
(
f
x
=
xft
f
x
fy )
−
y
vy
fyft
f 2x
fxfy
vx
f
=
xft
f
−
xfy
f 2y
vy
fyft
M v = −b.
(10.7)
If the matrix M is invertible (full rank), then the velocity can be
estimated by simply left multiplying by the inverse matrix:
v = −M−1b
(10.8)
The critical question then is, when is the matrix M invertible?
Generally speaking the matrix is rank deficient, and hence not
invertible, when the intensity variation in a local image neighbor-
hood varies only one-dimensionally (e.g., fx = 0 or fy = 0) or
zero-dimensionally (fx = 0 and fy = 0). These singularities are
sometimes referred to as the aperture and blank wall problem.
The motion at such points simply can not be estimated.
65
Motion estimation then reduces to computing, for each point in
space and time, the spatial/temporal derivatives fx, fy, and ft. Of
course the temporal derivative requires a minimum of two images,
and is typically estimated from between two and seven images.
The spatial/temporal derivatives are computed as follows. Given
a temporal sequence of N images, the spatial derivatives are com-
puted by first creating a temporally prefiltered image. The spatial
derivative in the horizontal direction fx is estimated by prefiltering
this image in the vertical y direction and differentiating in x. Simi-
larly, the spatial derivative in the vertical direction fy is estimated
by prefiltering in the horizontal x direction and differentiating in
y. Finally, the temporal derivative is estimated by temporally dif-
ferentiating the original N images, and prefiltering the result in
both the x and y directions. The choice of filters depends on the
image sequence length: an N tap pre/derivative filter pair is used
for an image sequence of length N (See Section 7).
10.2 Differential Stereo
Motion estimation involves determining, from a single stationary
camera, how much an object moves over time (its velocity). Stereo
estimation involves determining the displacement disparity of a
stationary object as it is imaged onto a pair of spatially offset
cameras. As illustrated in Figure 10.2, these problems are virtu-
ally identical: velocity (v) ≡ disparity (∆). Motion and stereo
estimation are often considered as separate problems. Motion is
thought of in a continuous (differential) framework, while stereo,
with its discrete pair of images, is thought of in terms of a dis-
crete matching problem. This dichotomy is unnecessary: stereo
V
estimation can be cast within a differential framework.
Stereo estimation typically involves a pair of cameras spatially
offset in the horizontal direction such that their optical axis remain
parallel (Figure 10.2). Denoting an image as f (x, y), the image
that is formed by translating the camera in a purely horizontal
direction is given by f (x + ∆(x, y), y). If a point in the world
(X, Y, Z) is imaged to the image position (x, y), then the shift
∆(x, y) is inversely proportional to the distance Z (i.e., nearby
objects have large disparities, relative to distant objects). Given
D
this, a stereo pair of images is denoted as:
Figure 10.2 Motion and
Stereo
fL(x + δ(x, y), y)
and
fR(x − δ(x,y),y),
(10.9)
where the disparity ∆ = 2δ. Our task is to determine, for each
point in the image, the disparity (δ) between the left and right
images. That is, to find the shift that brings the stereo pair back
66
into register. To this end, we write a quadratic error function to
be minimized:
E(δ(x, y)) = [fL(x + δ(x, y), y) − fR(x − δ(x,y),y)]2. (10.10)
In this form, solving for δ is non-trivial. We may simplify things
by expressing the image pair in terms of their truncated first-order
Taylor series expansion:
f (x + δ(x, y), y) = f (x, y) + δ(x, y)fx(x, y),
(10.11)
where fx(x, y) denotes the partial derivative of f with respect to
x. With this first-order approximation, the error function to be
minimized takes the form:
E(δ) = [(fL + δ(fL)x) − (fR − δ(fR)x)]2
= [(fL − fR) + δ(fL + fR)x]2,
(10.12)
where for notational convenience, the spatial parameters have been
dropped. Differentiating, setting the result equal to zero and solv-
ing for δ yields:
dE(δ)
= 2(f
dδ
L + fR)x[(fL − fR) + δ(fL + fR)x]
= 0
f
δ = − L − fR
(10.13)
(fL + fR)x
Stereo estimation then reduces to computing, for each point in the
image, spatial derivatives and the difference between the left and
right stereo pair (a crude derivative with respect to viewpoint).
Why, if motion and stereo estimation are similar, do the math-
ematical formulations look so different? Upon closer inspection
they are in fact quite similar. The above formulation amounts to
a constrained version of motion estimation. In particular, because
of the strictly horizontal shift of the camera pair, the disparity
was constrained along the horizontal direction. If we reconsider
the motion estimation formulation assuming motion only along
the horizontal direction, then the similarity of the formulations
becomes evident. Recall that in motion estimation the brightness
constancy assumption led to the following constraint:
∂f ∂x
∂f ∂y
∂f
+
+
= 0,
(10.14)
∂x ∂t
∂y ∂t
∂t
Constraining the motion along the vertical y direction to be zero
yields:
∂f ∂x
∂f
+
= 0,
(10.15)
∂x ∂t
∂t
67
where the partial derivative of the spatial parameter x with re-
spect to time correspond to the motion (speed) in the horizontal
direction:
fxvx + ft = 0.
(10.16)
Unlike before, this leads to a single constraint with a single un-
known which can be solved for directly:
f
v
t
x
= − .
(10.17)
fx
This solution now looks very similar to the solution for differential
stereo in Equation 10.13. In both solutions the numerator is a
derivative, in one case with respect to time (motion) and in the
other with respect to viewpoint (stereo). Also in both solutions,
the denominator is a spatial derivative. In the stereo case, the
denominator consists of the spatial derivative of the sum of the
left and right image pair. This may seem odd, but recall that
differentiation of a multi-dimensional function requires differenti-
ating along the desired dimension and prefiltering along all other
dimensions (in this case the viewpoint dimension).
In both the differential motion and stereo formulations there exists
singularities when the denominator (spatial derivative) is zero. As
with the earlier motion estimation this can be partially alleviated
by integrating the disparities over a local image neighborhood.
However, if the spatial derivative is zero over a large area, corre-
sponding to a surface in the world with no texture, then disparities
at these points simply can not be estimated.
68
11. Useful Tools
11-1 Expectation/
11.1 Expectation/Maximization
Maximization
The Expectation/Maximization (EM) algorithm simultaneously seg-
11-2 Principal
ments and fits data generated from multiple parametric models.
Component
For example, shown in Figure 11.1 are a collection of data points
Analysis
(x, y) generated from one of two linear models of the form:
11-3 Independent
y(i) = a1x(i) + b1 + n1(i) or y(i) = a2x(i) + b2 + n2(i),(11.1)
Component
Analysis
where the model parameters are a1, b1 and a2, b2, and the system
is modeled with additive noise n1(i) and n2(i).
If we are told the model parameters, then determining which data
point was generated by which model would be a simple matter of
choosing, for each data point i, the model k that minimizes the
error between the data and the model prediction:
rk(i) = |akx(i) + bk − y(i))|,
(11.2)
for k = 1, 2 in our current example. On the other hand, if we are
Figure 11.1 Data
from
told which data points were generated by which model, then esti-
two models
mating the model parameters reduces to solving, for each model
k, an over-constrained set of linear equations:
xk(1) 1
yk(1)
xk(2) 1 a
y
k
k(2)
. .
=
.
,
(11.3)
. . b
.
. . k
.
xk(n) 1
yk(n)
where the xk(i) and yk(i) all belong to model k. In either case,
knowing one piece of information (the model assignment or param-
eters) makes determining the other relatively easy. But, lacking
either piece of information makes this a considerably more difficult
estimation problem. The EM algorithm is an iterative two step
algorithm that estimates both the model assignment and param-
eters.
The “E-step” of EM assumes that the model parameters are known
(initially, the model parameters can be assigned random values)
and calculates the likelihood of each data point belonging to each
model. In so doing the model assignment is made in a “soft”
probabilistic fashion. That is, each data point is not explicitly
assigned a single model, instead each data point i is assigned a
69
probability of it belonging to each model k. For each model the
residual error is first computed as:
rk(i) = akx(i) + bk − y(i)
(11.4)
from which the likelihoods are calculated. We ask, what is the
likelihood of point i belonging to model k given the residual error.
For our two model example:
P (r
P (a
k(i)|ak, bk)P (ak, bk)
k, bk |rk(i)) =
P (rk(i))
P (r
=
k(i)|ak, bk)
, (11.5)
P (r1(i)|ak,bk) + P(r2(i)|ak,bk)
for k = 1, 2. The expansion of the conditional probability is from
Bayes rule: P (B|Ak) = P(Ak|B)P(B) . If we assume a Gaussian
P (A
l
l|B)P (B)
probability distribution, then the likelihood takes the form:
e−rk(i)2/σN
wk(i) =
(11.6)
e−r1(i)2/σN + e−r2(i)2/σN
where, σN is proportional to the amount of noise in the data, and
for each data point i,
k wk(i) = 1.
The “M-step” of EM takes the likelihood of each data point be-
longing to each model, and re-estimates the model parameters
using weighted least-squares. That is, the following weighted error
function on the model parameters is minimized:
Ek(ak, bk) =
wk(i)[akx(i) + bk − y(i)]2.
(11.7)
i
The intuition here is that each data point contributes to the esti-
mation of each model’s parameters in proportion to the belief that
it belongs to that particular model. This quadratic error function
is minimized by computing the partial derivatives with respect to
the model parameters, setting the result equal to zero and solving
for the model parameters. Differentiating:
∂Ek(ak, bk) =
2w
∂a
k(i)x(i)[ak x(i) + bk − y(i)]
k
i
∂Ek(ak, bk) =
2w
∂b
k(i)[ak x(i) + bk − y(i)],
(11.8)
k
i
setting both equal to zero yields the following set of linear equa-
tions:
ak
wk(i)x(i)2 + bk
wk(i)x(i) =
wk(i)x(i)y(i) (11.9)
i
i
i
ak
wk(i)x(i) + bk
wk(i) =
wk(i)y(i). (11.10)
i
i
i
70
Rewriting in matrix form:
i wk(i)x(i)2
i wk(i)x(i)
ak
=
i wk(i)x(i)y(i)
i wk(i)x(i)
i wk(i)
bk
i w(i)y(i)
Axk = b
xk = A−1b,
(11.11)
yields a weighted least squares solution for the model parameters.
Note that this solution is identical to solving the set of linear
equations in Equation (11.3) using weighted least-squares.
The EM algorithm iteratively executes the “E” and “M” step,
repeatedly estimating and refining the model assignments and pa-
rameters. Shown in Figure 11.2 are several iterations of EM ap-
plied to fitting data generated from two linear models. Initially,
the model parameters are randomly assigned, and after six itera-
tions, the algorithm converges to a solution. Beyond the current
scope are proofs on the convergence and rate of convergence of
EM. At the end of this chapter is a Matlab implementation of
EM for fitting multiple linear models to two-dimensional data.
11.2 Principal Component Analysis
11.3 Independent Component Analysis
Figure 11.2 Six
iterations of EM
71
%%% THE EM ALGORITHM
clear;
NumModel
= 2;
NumData
= 64;
Sigma
= 0.1; % IN E-STEP
Noise
= 0.1; % IN DATA
NIterate
= 10;
%%% MAKE SYNTHETIC DATA
A
= 2*rand(1,NumModel) - 1;
B
= 2*rand(1,NumModel) - 1;
X
= [];
Y
= [];
for i = 1 : NumModel
x = 2*rand(1,NumData) - 1;
y = A(i) * x + B(i) + Noise*(rand(1,NumData)-0.5);
X = [X x];
Y = [Y y];
end
%%% INITIALIZE MODEL
a
= 2*rand(1,NumModel) - 1;
b
= 2*rand(1,NumModel) - 1;
for j = 1 : NIterate
%%% E-STEP
for i = 1 : NumModel
r(i,:) = a(i)*X + b(i) - Y;
end
den
= sum( exp( -(r.^2)/Sigma^2 ) );
for i = 1 : NumModel
w(i,:)
= exp( (-(r(i,:).^2)/Sigma^2) ) ./ (den+eps);
end
%%% M-STEP
for i = 1 : NumModel
M = [ sum(w(i,:).*X.^2) sum(w(i,:).*X) ; sum(w(i,:).*X) sum(w(i,:)) ];
V = [ sum(w(i,:).*X.*Y) ; sum(w(i,:).*Y) ];
U = pinv(M) * V;
a(i) = U(1);
b(i) = U(2);
end
%%% SHOW MODEL FIT AND ASSIGNMENT
xc
= [-1 : 0.1 : 1];
subplot(2,1,1); cla; hold on;
plot( X, Y, ’bo’ );
for i = 1 : NumModel
yc = a(i)*xc + b(i);
plot( xc, yc, ’r’ );
end
hold off;
subplot(2,1,2); cla; imagesc(w); colormap gray; pause(1);
end
72
